Linear and General Functions of Random Variables

okay this lecture is going to talk about linear functions of random variables and general functions of random variables in OEE 380 and in the early part of this course we learned about you know single random variables and now joint probability functions and now we're going to look at functions of random variables that are also random variable so for example here we have Y which is a constant C one times another random variable X 1 plus C 2 times another random variable X 2 and so on sorry I got to my sound off on my computer there we go and so Y is going to be this linear combination of these different random variables these X's now a couple of things that we need to learn about which I think are actually in an earlier chapter but if we have before I get going I need to you know these things so suppose a and B are constants we call c1 or c2 it really doesn't matter the expected value of a constant is just the constant and that should make sense to you like I always do this in class I'm going to do it here if I put on my hand the number 4 I don't know I always choose 4 but I do and I close my hand and I say to you what's the probability that there's a 4 on my hand well it's a 1 I just it's there so 4 times the probability that I have a 4 in my hand is just a 4 so the expected value the constant is always a constant the expected value of a constant times a random variable plus another constant turns out to be that constant times the expected value of x plus B I could have also written it this way it would have been the expected value of a times X plus the expected value of this constant B which works out to be this a time's the expected value of X plus B because we just learned that the expected value of a constant is a constant now I'm not going to go into the proof of this but this is kind of interesting the variance of a linear function here in ax is just a squared the constant squared times the variance of X the variance of a constant maybe I could put that up here the variance of a constant is always equal to zero why is that because this for in my hand with a probability of one there is no variability there so that's why the variance of a constant is zero and that's what this piece of this component is here now if x and y are independent and this is a pretty big F then the variance of a linear combination of two random variables here x and y or it could be x1 and y1 x2 whatever is just going to be a squared times the variance of X plus B squared times the variance of Y and so on I'm I can keep going I mean going going up with this and that's what I'm going to do actually on the next slide so suppose that I do have these X's and the book I don't know why but they subscript it by P I would have like K or N or I it doesn't matter but and we create we have these this linear function in X 1 X 2 XP with these different constants and the constants could be the same value or different values these X's can come from the same distribution or from different distributions then this is how I'm going to get the expected value of y per the previous slide so it'll be c1 times the expected value of x1 c2 times the expected value of x2 and so on now if my random variables x and y are independent then the variance of this linear combination becomes per the last slide the constant squared times the variance of x1 plus that second constant squared times the variance of X 2 and so on dot dot dot plus CP squared times the variance of X P and we are going to use this equation on the next slide now suppose that i have this semiconductor product and it has three independent layers so i'd like to draw pictures as you know so we have the semiconductor and it has three layers those me layer one they don't necessarily have to be exactly the same bit but i'm going to call this layer I'm going to call X is the thickness of layer 1 okay so that's gonna be a random variable and I think I'll use well let subscript them let's just make them x1 is the thickness of layer 1 and then a couple of different colors here let's let X 2 is the thickness of layer 2 and lastly I mean pink cuz it's working x3 is the thickness of layer 3 now the problem here is that they don't really they're not asking me a question about X 1 X 2 X 3 specifically what they want to know is what is the standard deviation of the finished product well the finished product is going to be Y right here so Y is the thickness of the finished product or the finished semiconductor so Y in terms of X 1 X 2 and X 3 is just the sum of them so Y is equal to X 1 it's a random variable okay plus X 2 plus X 3 and x3 here okay what's the question about them what they want to know is what's the standard deviation of the finished product so what they want is they want me to find Sigma sub y okay which I'm going to get by taking the variance of Y and the square root of that so which is the same thing as me saying take the square root of the variance of Y okay so that's going to be pretty easy to do the variance of Y so I'll just continue on here why not the variance of Y I'm just gonna plug in that would be the variance of this X 1 plus X 2 plus X 3 and I guess I'll continue to use my colors here oops thanks three okay now we know since these are our linear combination and the key word here is independent so because they're independent that means that I have see I have this invisible by the way I have an invisible one here an invisible one here an invisible one here if they're independent that means that again this mu square root that this equation here this is just going to be the variance of X 1 R 1 squared times the variance of X 1 is an invisible one there right plus my invisible 1 squared times the variance of X 2 plus my invisible 1 squared times the variance of X 3 and now they've given me the variances of each one of these so up here it says the variance of the thickness of each layer is 25 40 and 30 so they told me that the variance of X 1 is equal to 25 nanometers and at the bare x2 oh you know what that would be 9 meters squared sorry about that there we go and this will be 40 nanometers squared finally the variance of X 3 that's gonna be 30 meters squared forgot to square that term when I was making it so now all I'm gonna do is I'm just gonna add up all of these variances and then take the square root and I have the answer to this to this problem so we have the square root of the variant we have 25 plus 40 plus 30 and last time I checked that's going to be the square root of 95 and so the answer to this problem I took a very long time to do a pretty simple problem but I kind of want to do all of the steps here the square root of 95 is nine point seven four six eight and that would be in nine leaders so that is the answer to that one pretty easy to to do that one now you've actually seen this before and I II III eighty and let me show you where you've seen this remember when we're taking the sample average from a population we take a bunch of we take a bunch of values from the population we add them up divide by the number in my sample that's what we call the sample mean and the sample mean is given here now X itself is a random variable and it has a mean of mu and a variance of Sigma but then X bar is also a random variable it has its own expectation what else which is the same as the expectation of the underlying distribution but the variance of X bar is not quite the same and you might remember from 380 the way I teach it in 380 is as follows I like to show two different urns so how do we get x-bar so let me do it up here so imagine that I have an urn here and my urn is full of my random variable X single values of it and we know that my expected value of x is equal to mu and we know that my variance of X is equal to Sigma squared so that's my urn it's kind of cramped doesn't really matter what the distribution is so notice I didn't draw a distribution now suppose that we reach into this urn and we grab a sample of size n so we're going to take X 1 X 2 X 3 we're going to take a bunch of them so I'm going to take a sample of size n so we'll take what my first draw my second draw and there's gonna be one of those and they're all capital letters because they're all random variables when I reach in and actually record the value than their lowercase values right I'm going to take a sample of N and then I'm going to create a sample mean here and I'll call it X bar and then I'm going to take each one of those X's and I'm going to add them up and divide by N and that will be X bar well for every time I go in here for each sample of n I'm going to create a separate value of x bar and I'm going to plunk that in here so this is going to be an urn of all my random variables X bar well it turns out that the expected value of all of the values that are in this urn just happens to be exactly the mean of the underlying distribution that I got my sample from in the first place so these are exactly the same so that's easy to convenient but the variance in this urn of X bar that is equal to Sigma squared which is this guy right here but divided by N I'm going to highlight that Sigma squared in orange and that's how I teach it in IE 380 and we just asked you to accept it that's true let me show you why it is true because we just learned why it is true if each one of these are independent so we're assuming that these are independent then we know that this is going to be a linear function of random variables right so this is the equation right here that I use to create each one of those X bars now it's easy to show that the expected value of x bar is Mew so let me just take this equation right here and I'm going to say take the expected value of X bar and they're just going to write this out so this is the same as saying 1 over n times X 1 plus 1 over n times X 2 plus all the way up to 1 over n times X n well what we just learned is that the expected value of a linear function of these random variables I have to put an expected expectation operator here well this is the same thing as 1 over n times the expected value of X 1 plus 1 over n times the expected value of x 2 and so on X n and we know that each one of these expected values is just equal to MU so the expected value of x bar is going to be 1 over n times mu for this term here plus 1 over n times mu this term here plus 1 over n times this term here well I have n of these terms so I can I can get it up my terms here and I have 1 over N times and I have of these muse the ends canceled out and we had new so what I just was able to show you is why this is true so that's a little demonstration there now the variance is similar and let me do that under that a different color just to have it pop out a little bit let's see if this works out so based upon what we know about the variance of a linear function of random variables so the variance of X bar is I'm going to take the variance of this equation right here so I'm just going to expand this equation just like I did it here so X bar will be x1 plus x2 plus x3 oops plus xn all over N that's just this right here then I could also expand it this way I can say well this is the same thing as 1 over N times x1 plus 1 over n times x2 plus 1 over N times X N and so what we just learned is if these are all independent then I can write those as the variance of 1 over N times x1 plus the variance of 1 over N times x2 and so on whoops variance I'm right there we go and now for the variance of each of these I pull the constant out and square it so now I have 1 over N squared times the variance of x1 plus 1 over N squared times the variance of X 2 plus 1 over N squared times the very and each one of these variances though has this value right there and how many of these terms do I have I have n of them so the variance of X bar is equal to 1 over N squared and then I have n times Sigma squared which is from this distribution here and that works out to be Sigma squared over n so what I was just able to do is demonstrate for you that why this equation has that where that equation is that equation so these are values that we've used before it's just that now I'm showing you why why those equations are true now problem 9 gives us an example and I and I like problem 9 which is the next one here because what it does it can show you how you can make a mistake and what the difference is between multiplying a constant times the random variable and a linear function of random variables so here what let me just give you the short version here this is a study about how long it takes to do a particular knee surgery okay now they're telling me that the knee surgery mean time is about 118 minutes about two hours has a standard deviation of 15 minutes for this ACL reconstruction surgery okay what they want to know is if we've got this at the hospital and they need to schedule 10 surgeries now what they don't say here is that the surgeries are all going to be going basically back-to-back okay which is probably not true in a big Hospital you would be able to have you know more than one operating room but we're going to assume that we just have one operating room here and we have to schedule these surgeries so what they want to know is what's the mean and the variance of the total time to complete these surgeries so I'm going to write down in blue cuz I like this pen a lot I'm going to write that down at is the time of a surgery how long it takes now we can think of that as my urn of X's right so I have a bunch of X's here okay and in this urn we know that the expected value of any one of those surgeries is equal to one hundred and eighteen minutes because they told me that and they also told me that the standard deviation of any particular surgery is equal to fifteen minutes so that's my urn now the thing is I need ten of these so what I can do is I can subscript them so if I grab ten different surgeries because not every surgery is going to be exactly one hundred and eighteen minutes right so maybe my knee surgery is you know an hour and 10 minutes maybe somebody else's knee surgery is closer to three hours because there's some complications so each one of these is a random variable with a mean of 118 and a standard deviation of 15 that means that if I have to schedule 10 of these things that I have 10 here so I'm going to let Y be is the duration of 10 surgeries so that means that Y is going to equal 1 2 3 all the way up to 10 so that means so the first person surgery is a random variable then the second person surgery is a random variable up to the tenth person surgery is a random variable now I want to pause for a second that was a mistake that I actually made a number of years ago so I read the question fast was I did so this is wrong I said Oh Y is equal to 10 times at now that is not the same you're gonna end up with the same mean here if I take the expected value of the Green Y and the expected value of the red Y they're going to end up being the same value but the variance is going to be option here's why this says that I have 10 individuals with individual random variables here so this is a linear combination of ten random variables that all just happen to come from the same distribution this says that Y is equal to ten times just one of those random variables so this is wrong so I wanted to point that out this is not the same as this okay this is a general function of a random variable which we're going to be getting to so now what we want to know is what is the mean and variance of the total time to complete these surgeries so what do they want me to find they want me to find the expected value of y and the expected value of y is just going to be the expected value of each one of these X's so that's the expected value of x one plus the expected value of x two plus the expected value of x ten and each one of those is this mean of 118 so we'd have 118 plus one 18 and there's ten of them so that's gonna be one thousand one hundred and eighty minutes so that's an easy one to do the variance of Y because there's also another keyword in here right where is that keyword right here independent so I can do this because these X's are independent now the variance of Y that's going to be member there's invisible ones up here right so we have one squared times the variance of x one plus one squared times the variance of X two plus one squared times the variance of X ten and each one of these variances is going to be fifteen squared because that's Sigma squared and we also that Sigma is 15 so Sigma squared must be 15 minutes squared okay so that would be 15 squared is 225 so we're going to add 225 plus 225 plus 225 and there's ten of those so we'll have two two five Oh minutes squared that's the variance of Y so what if I said this if I said well what if the problem up here said what is the probability let me just add that just kind of a value-added question here what is the probability that what is the probability that a total time of ten surgeries is let's say less than just making something up here mm I'm sorry well o less than I'll say hang on I'm gonna make this more interesting than it actually is let's say less than 18 hours so what do they want to know they want to know what is the probability that this random variable Y is less than 18 hours well how many minutes is 18 hours 18 hours times 60 minutes in an hour so this is the same thing as saying what's the probability that Y is less than 1080 minutes what is the distribution of Y well we know from the central limit theorem from I a 380 that when I take the sum of or x-bar or add these things up if if Y if each one of these is normally distributed then Y is normally distributed well they've assumed that right so the sum of normally distributed random variables per the central limit theorem is also normally distributed so we happen to know that my random variable Y has a normal distribution with a mean of eleven hundred and eighty minutes and a variance of 2,250 minutes square so this probability is very easy to solve that's just going to be normal CDF and I'm going to go from negative infinity up to 1080 minutes for a mean of eleven hundred and eighty minutes and a standard deviation equal to and then we'd have to do the square root of two to five oh and I'd actually didn't do this before I got here so well I've been here all day I'm in my house but okay so if I do this probability what do we get just curious I'm gonna go from negative infinity which I like to do is negative ten to the 99 up to 108 oh for a mean of one one eight zero and a standard deviation equal to the square root of twenty to fifty and that works out to be point zero one seven five that would be more likely a question that I would put on a quiz or an exam because it's an application all righty uh next next there are general functions of random variables so what I said here was this is what we call a linear function of some random variables here Y is a linear function this is an example of where Y is just a general function of a random variable and so that's where we're going to go then this is wrong for this problem but it is an example of the next topic okay so general functions of random variables that is tada right there I'm going to separate this out by the street and by continuous so for suppose I have X and it's a discrete random variable really doesn't matter what its probability distribution is it could be something known it could be something unknown it could be something empirical something with a name etc etc so this Y suppose this Y is just some function of X ok so that looks fancy but it's just a way of saying that Y is a function of X how do I get the probability mass function of Y and I have basically a two-step procedure here that I like to do I don't know that the book actually puts it this way but I like to do it this way so I'm gonna give you two examples of this the first one is a pretty easy example suppose that I have ax it's a geometric random variable and suppose it has this probability distribution so just so you know a reminder that I use this notation I use a lowercase P so if X has a geometric distribution we know that a geometric random variable is the trial upon which the first success occurs and this is a Bernoulli trial success probability this is a Bernoulli trial trial failure probability and so the first success can happen earliest on the first trial so kind of a reminder is something from Chapter three what if for some reason I have a new random variable Y and I just want to take each one of these X's and square it this is what on the previous slide would be H of X it's just a function in X okay so according to the previous slide the first thing I want to do is identify what my range of Y is going to be right so that's pretty easy to do for a discrete random variable step one going to take my values of X and I'm going to figure out what my corresponding values of Y are going to be so when X is 1 Y is going to be 1 squared when X is 2 y is going to be 2 squared and so on 3 this is going to be 3 squared is 9 okay so now I know what my range on Y is that's that first step so range of Y is going to be y equals 1 4 9 16 and it goes on with that pattern the second step says that we are supposed to figure out what each of the values when we go back to one slide here go back to slide 41 I want to say this a little differently we just did step one step two says some of the PMF values of x that correspond to each Y to obtain P of Y I wouldn't say sum I would say determine sometimes you can use the sum if you want to figure out what the CDF is first so please replace the word son with determine I'm not very happy with that I caught that in a number of years actually ok so back to our problem here so we are pretty weak this is just a substitution so it's pretty easy to do so I can express my random variable X just by taking the square root here of Y right easy peasy now now if you just do a substitution so my probability mass function of Y I'm just wherever I see an X I'm going to put the square root of Y and wherever I see an X here I'm going to put the range of Y's this is a really easy one so I'm gonna take my Bernoulli success probability my Bernoulli failure probability and then stead of X minus 1 I'm just gonna put the square root of Y and then minus 1 and that's just the square root of y there and that's gonna be good for y going from 1 y equals 1 4 9 16 etc so that's pretty easy right there almost too easy problem 11 is when we have a discrete random variable that has an empirical distribution meaning that it doesn't have a name it's unique to a particular situation so here suppose that I have my random variable X and it happens to have this probability mass function so when X is minus 1 we have a probability of 0.6 when it's 1.3 and when it's 2 its 0.1 notice that these all add up to 1 is just a regular normal probability mass function now suppose I define my random variable Y to be 20 times x squared this does determine the probability mass function and the CDF of Y now let's follow my steps step 1 determine the range of Y this is going to be pretty easy to do also because if I look at my X and I looked at my Y what do we have when X is minus 1 Y is going to be 20 x minus 1 squared so that's just going to be 20 y is equal to 20 the next value of x when X is 1 Y is equal to 20 times 1 squared hope also 20 and then finally when X is equal to 2 y will be 20 times 2 squared so that's going to be 80 so here I do not have a one-to-one correspondence here X can be 3 values but Y can only be 2 so Y is permitted to be 20 or 80 that's it step two says now we're going to determine the individual probability mass function for y so the probability that Y is equal to 20 is going to be the probability that X is equal to do it this way is the probability that x is equal to minus 1 or when x is equal to 1 so y equals 20 that probability is going to be 0.6 plus 0.3 to 0 point I never get myself enough room and then what is the probability Y is equal to 80 well Y is 80 only when X is equal to 2 so that would be zero point one zero so that's the probability mass function of Y if I wanted to show the CDF of Y well that's pretty easy too we know that whenever Y is less than 20 but the CDF will be zero and we also know that whenever y is bigger than or equal to 80 that the CDF has to be one so there's some intermediate value in here where Y is not quite 80 but bigger than or equal to 20 and that would be 0.9 zero this usually confuses people because CDF is not an easy function to remember but this is not the probability that Y is between these values it says whenever I plug in a y here this is the probability so let's do it suppose I said what's the probability that Y is less than or equal to 30 well it can't be 30 but it can be 20 so the probability that Y is less than or equal to 30 is just the probability that Y is less than or equal to 20 so for this range on Y that it's this value of 0.9 zero this is a little bit of a refresher I suppose okay now the more complicated one which I think is a lot more fun quite frankly is when I have continuous random variables and I get to do a little bit of calculus so I love doing calculus this would be a good time by the way to pause take a break go have some dinner because I'm kind of powering through this lecture here general functions of random variables for continuous random variables now same idea I've got a new random variable Y and it is some function of random variable X now how do I get the PDF of why this takes a little bit more work actually but as I said I think it's a little bit more fun I'm gonna find the range of Y just like I did with discrete but then I'm going to create the CDF of Y by plugging in what my Y is equal to so where it says here my CDF of Y is the probability y is less than or equal to this little value right on a subscript this guy here well Y it is just going to be some function in X so that just I'm just gonna plug in what the equation is for y less than or equal to Y and then I'm gonna solve it in terms of X and I'll be able to get the CDF of Y so then I'm going to take the first derivative of Y so I go D dy and we know that that gets me the probability mass function for y so when I take the first derivative of a CDF I get the PDF will never do the integral over the PDF and I get the CDF okay so let's just jump one into one here because I like this one quite a bit suppose I have and then I'll do one that actually where we where it's an actual application this is just a general kind of a nasty problem right here I got this continuous random variable here in X and it actually happens to have the triangular distribution you might remember we covered that when I was doing continuous random variables and since find the probability distribution of Y which is maybe two times a random variable from this distribution plus four so just for fun let's take a look at what this distribution would look like if I were to plot what X looks like the PDF f of X says that X goes from zero to four right here and so I'm just gonna plot the function so when and this is a you should recognize this as y equals MX plus B then B being the y-intercept well the y-intercept is zero so I can just see I'll just plug in two values so when X is zero then this function is zero when X is 4 then this function is 1/2 so this is a plot of the PDF of my random variable X you should recognize this as a triangular PDF that we learned about earlier okay we know that the area under it has to be 1 which is easy 1/2 base times height all right great so the point is is I'm going to take an X from here and then I'm going to multiply it by 2 and add 4 and that's what's going to get me Y so what I want is I'm trying to find over here what do they want they want me to find or determine my PDF of Y ok so let's follow the steps on the previous page step 1 says find the range of Y well here I'm not going to make a table of x and y it's just it's pretty easy to see so I well I can do it that way let's do it this way so my low X is equal to 0 so that means that my lowest value of y is going to be 2 times 0 plus 4 equals 4 and then my highest value of ax is 4 which means my highest value of y will be 2 times 4 plus 4 equals 12 so in step one we have determined that Y is gonna go from 4 to 12 that's the range of Y step 2 says we're gonna find the CDF of Y so we're gonna start out this way we say what is the CDF of Y we're going to do a little substitution so I'm gonna do this painstakingly slow because you don't want to miss a step so this is the probability that my random variable Y is less than or equal to Y that is just a definition of what a CV at this but why has this formula here so I am going to plug in this formula so that's the same thing as 2x plus four listen or equal to Y now I do have the probability distribution for X okay so I'm going to get X by itself on the left hand side of this inequality and that's gonna make my life a lot easier so if I do that this is like saying what's the probability that X is less than or equal to Y minus four over two so what is this this is just the CDF of X which we haven't figured out yet evaluate it at a little X that is equal to prep Y minus four over two so what is this if I took this little X guy right here and I said I'm just gonna call you y- floor like I've renamed it over to then this integral right here is what capital f of X is evaluated at that point when I do that that's gonna equal my CDF for y so to get this value all I need to do is integrate from my lowest value of x we know how did this is the general form right we go low x to general x over my PDF of X DX and when I do this I'm gonna go from 0 to oh I'm sorry hang on yeah if I wanted a general axe oh hang on a second but this general x value now is going to be this value here so I'm gonna go up to an X that we're renaming Y minus 4 over 2 of this PDF D X ok now I'm gonna save you a little bit of time but I want you to do this yourself okay this will equal do the integral yourself all right because you will have to do it on a quiz you will have to do it on a test do it yourself and prove to yourself that when you do this integration you are going to end up with the following you will get Y squared minus 8y plus 16 all over 64 let me remind you what this is this is going to be capital F of Y I just brought this over here so this is going to be the the functional portion of F Y of Y this is going to be the piece for when Y is between 12 and 4 okay this is not what we were asked for what we were asked to find was the PDF of Y right there and that's what we're going to do in step three so step three says we're gonna take the first derivative with respect to Y of this function again you do that you do that integral I need you to do that because you'll have to do it on a quiz or a test so when we take the first derivative of this what do we get well we're going to get 2y minus 8 over 64 which if we wanted to simplify it we can divide the numerator and the denominator by 2 so we would have Y minus 4 over 32 and that that's the answer let me write it down over here so prettier would be this way here is my PDF of 1 that will be Y minus 4 over 32 and that will be good for y between 4 and 12 that is the answer once you have the answer though now what could we do we could say well what's the expected value of y well that's easy to get actually the expected value of y and we do it over there so it would be fairly easy to find suppose I wanted the expected value of y well I could I could do it the hard way which would be to integrate from four to twelve over this function right here or I could just do this substitution and write the expected value of 2x plus 4 and that would be 2 times the expected value of x plus 4 and then well I guess it's the hard way one way or the other so that would be 2 times and I can get the expected value of x here by integrating from 0 to 4 oh we also know the expected value of a triangular distribution right we actually do know that but let me just show you this so this would be 2 times X times X over 8 DX and then plus 4 anyway you can get the expected value of y a couple of different ways I get the variance of Y a couple of different ways and I can ask questions about the probability of Y so this is how you would find the PDF of a continuous random variable Y that is some other function of another random variable so let me do another example and this actually came from I think an exam that I did a couple of years ago so I'm calling it problem 13 a B and C and after this problem I'll stop and then I'll do the next lecture for the week problem 13 a and 13 B we have a tire shop and get discounts to customers according to how long I've waited so the longer they wait the better the discount they get now it says based upon you know some other I guess analysis the time it takes for a custom the time that they're gonna wait we're gonna call time T capital T here this is a random variable this is not like the big t that we use in the poisson random variable and when somebody has determined this is exponentially distributed it hasn't mean equal to 1.5 hours when the car is ready they're gonna receive a discount D which is also a random variable and they're gonna say okay as long as you've waited we're going to multiply that by um 10 okay so show the equation for D as a function of T now this is an example where we have one function we have a random variable D that is a function of another random variable it's not the sum of a bunch of them it's just a function of it so according to these words here it says that my discount which is going to be in terms of dollars is going to be equal to $10 times however long I wait which is another random variable T so that is the answer to a very very easy based upon the answer above for Part B what is my expected discount now it says circle the correct answer on the answer sheet because I had it as a multiple choice on that previous test but what's the expected discount where they want me to do they want me to find the expected value of the discount well I can just plug that in that's the expected value of 10 times T which we know is 10 times the expected value of T and we told me up here that T was an exponential random variable um so before I finish this up let me write out something else we know from the problem it told me that the probability density function of the time that we wait is exponential with a mean equal to 1.5 hours well the expected value of T is 1.5 hours which means that lambda is equal to 1 over 1.5 I'll just leave it like that so we have 1 over 1.5 e to the minus T over 1 point and that would be good for T bigger than zero I could also have written it 1.5 is also three-halves I think I just want to show you this oh so we could have written it this way sometimes it's easier so one over three-halves just a little sidebar here is that 1 over 1.5 is the same as 1 over 3 halves so that just equals 2/3 so we could have written it this way 2/3 e to the minus 2t over 3 for T bigger than 0 I just wanted to just show you if I'd done it this way you would have been confused probably so ok but what is the expected discount coming back here 10 times the expected value of T the expected value of T is 1.5 hours so we will have 10 times 1.5 but we know that the units for my discount are going to be in dollars so my expected discount is going to be $15 so that's the answer for Part B now Part C Part C this is the same wording up here so there's nothing nothing new here develop the probability density function for D comma the amount of discount they receive show your work use proper notation and units and box your answer no box no credit this is like if it was on an exam cuz it was on an exam so what are they asking me to do here and these like pink color again find the probability density function for the discount okay that's what they want me to find so I'm gonna have to follow those steps we know that D is equal to 10 times T so following my original example step 1 says find the range on D well we know that T is bigger than 0 so T would equal to let me do it this way we can express T as d over 10 so if T is bigger than 0 then D over 10 is going to be bigger than 0 which means that D is bigger than 0 so not a very interesting change in range here D is going to have the same range as teeth step 2 says determine the CDF of D so what is that that's the probability that my discount is less than or equal to that lower case D I'm going to substitute in what my discount is we know that my discount is 10 times T then less than or equal to D I'm going to get it in terms of something that I know the distribution of so this is saying the probability that T is less than or equal to D over 10 so let's pause for a second if we know that T is exponentially distributed right that's what it's set up here so off to the side here let's just go ahead and show the distribution of T I wrote it before but we know that the distribution of T is 1 over 1.5 e to the minus 1.5 whoops 1 over 1.5 there and that's for T bigger than 0 well this is exponential we also know that the CDF of T is 1 minus e to the minus T over 1.5 for T bigger than 0 well that's exactly what this is right here this is the definition of right here this is exactly the definition of changing my colors here capital F of T but evaluated at a lowercase T of D over 10 so we just have to plug D over 10 into the CDF of and we have the CDF of D so now that would be one - and wherever I see the T I'm going to put D over 10 e to the minus D over 10 over 1.5 and if I make that a little bit nicer that would be one minus e to the minus D over 15 and that will be for D now but we're looking this is for D bigger than zero step 3 I'm going to take the first derivative shouldn't use the D huh with respect to D D D D you get the idea right of capital F of D so I'll write it this way that'll be or just F prime D of D so I'm gonna take the first derivative of this thing here and that's gonna end up being D over 15 right times a minus one might be this this and then this make the whole thing here minus D minus one and then e to the minus D over 15 and when I gather all of those terms it's just going to be whoops no D here sorry that's just a 1 that's 1 15 apologies there we go so now we're going to get 1 over 15 e to the minus D over 15 and that's for D bigger than zero and that is equal to that so the answer here not the prettiest that is the probability density function for my random variable D oh gosh sorry this is a capital D so sorry the lowercase D here there we go so it's actually not that difficult but this is why it's so important that you need to recognize these different distributions with names and you know I have these on your note sheet for sure for the next for the next exam because I will be using known distributions for Chapter five kinetics okay all I got way out of the way sorry about that okay on to the next lecture

Channel: Dr C

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Id: CGN5hsAELJY

Length: 56min 29sec (3389 seconds)

Published: Sat Mar 14 2020