Joint Probability Distributions for Continuous Random Variables - Worked Example

in this presentation we're going to look at probability distributions and in this particular case what we are looking at is a continuous probability just joint a continuous probability distribution my pen is all over the shop joint a continuous joint to distribution to continuous random variables essentially so and a certain factory produces two kinds of products in any given day widgets and gizmos okay now let these add two kinds of products be represented by the random variables x and y so X or widgets Y for gizmos given that the joint probability density function of these variables is given by such and such what we're asked to do is answer these questions okay so that's the joint probability distribution that's quite important okay so just as sort of for the sake of clarity I'll write it down here at the bottom again so it is 2/3 of X plus 2y for X less than or equal to 1 and y less than or equal to at between 0 and 1 that's a little bit of a shot there but bring that back in the shot and 0 elsewhere okay okay so what were asked to do is find the marginal PDF that's the probability density function of X and similarly do the same thing for y PDF is probability density function so how do we do that well what we do is so map back find the marginal PDF of X so we're going to call this G of X just to sort of not be using same letters all the time this is we're just going to call it G of X that's a G there so G of X is the integral from minus infinity to infinity this is just general expression of X Y D Y ok uh what I'll do here is I'll actually just go to the next page and what I'll do here is we're going to call this H of Y okay this is the match of the PDF of Y H of Y is the from minus infinity it is the integral of the joint PDF with respect to X ok sorry I just go to sort of highlight that fact here the G that the probability that's the virtual PDF of X is the integral of the joint distribution with respect to Y the PDF of Y is a joint is a integral of the joint distribution with respect to X just watch out for that it's very easy you can get them mixed up that's very important so anyway and let's see what's up all right let's go down here make some space so let's actually sort of calculate this out and what we're going to do here is I just got to restate the air limits because we don't have to go from infinity to minus infinity we can go from 0 to 1 because we're told as in the question okay so this is the integral of 2/3 x plus 2y dy okay and that is equal to 2 over 3 so what we're integrating here is X we would just say wait we separate out the two thirds separately ok we'll just take that out with our equation so what we're doing here is integrating X with respect to Y and then two Y with respect to Y and that is XY plus 2 y squared over 2 okay and integrated from 0 to 1 but the 2 x squared over 2 is essentially cancels out so 2 x squared over 2 there we go so that's what we have to do there and we let y equal to 1 that's actually sort of bit clear about this we're integrating with respect to Y not with respect to X so just be a little bit clearer about that and so it is 2/3 of x times 1 so that's X plus 1 squared that's 1 okay minus 2/3 at 0 minus 0 or 0 plus 0 okay so I know so trying out the cough there so the answer essentially is that Oh hits two thirds of X plus one now same thing for y h y equals two thirds let's just take out that a constant term and also we're evaluating from zero to one so it is not it is with respect to X where this integral is being calculated with respect to X it is let's do it again two thirds so that's been taken out of X plus 2y DX okay and that is equal to two thirds of x squared over two plus two X Y from zero to one okay calculating that out we will get two thirds of oh by the way sorry let's be absolutely clear crystal clear here again this is X equal to 1 X equal to zero okay and when we were dealing with like two different variables we're integrating with respect to two different variables just might be a little bit high it might be just a smart thing to do just be vet much more clear about what you're integrate there just for your own sake so that's 1/2 plus 2y minus 2/3 0 plus 0 so we just disregard that second part all together that just cancels it to 0 so the answer is two over three are at times 1/2 R plus 2y you could sort of like a go a little bit further and state it as 2/3 up 1/2 is 1 third plus four over three why something like that I'll tell you what we leave it as it is you can sort of like a a bring the cancel out of two thirds into it so what we're going to do now is find the probability that X is less than equal to half and y is less than equal to 1/2 so what we got to do here is two integrals okay so that is how do we calculate that that is a double integral so and both with respect to these limits okay not point five nine point five the X 2/3 of X plus 2y DX dy okay now I'll just be clear about this this limit here well I'm just started doing it in the order presented it shouldn't really matter but we're going to put a differentiate with respect to X first okay so that's an integral with respect to X and that's the limit corresponding to that value there and then the other than what we will do is differentiate that with respect to Y our outcome okay and this half here is from Y is less than equal to half that's let's be very clear about that okay so this at not 0.5 here is coming from here this not 0.5 here is coming from X less than equal to 0.5 so just be very clear careful about that just in case you get different numbers and ok so I just go do this first two double integral okay and I just sort of script out some of my work not point five okay so what we'll do is we'll do the inner part first okay and that is a two-thirds of x squared over two plus 2xy okay evaluation from null point five to zero okay so this is just what the this is just what's inside that red area there okay and that would work out to be not point oops gone too far at not 0.5 squared over 2 plus a 2 times X it is just simply 1 so simply Y okay and zeros and zeros the zero terms cancel out so minus 0 times 0 okay we can just disregard them but don't forget the 2/3 on the outside ok so let's actually sort of try and simplify that that is a not 0.5 squared over 3 plus 2 or 2 y over 3 actually not 0.5 squared is 1/4 so this is 1 over 12 this term the entire thing here is 1 over 12 okay so let's bring that back to the other part so what we're going to do now is integrate it with respect to Y okay so 1 over 12 plus 2y over 3 DUI and let's be clear about this that there is the inner part form of the solution to the inner part up there okay so let's keep going at this and the answer is y over 12 and plus two y squared over six evaluate from nine point five to zero so essentially we're just evaluating it at not 0.5 that would work out to be not 0.5 over 12 plus 2 times 1 over 4 over 6 okay so little bit of arithmetic to sort of sort out here and so 1 over 24 plus a2 times 1/4 is let's just said it's not 1/2 over 1/2 over 6 is a 2 over 24 okay so that it works out to be 3 over 24 and that actually works out to be 1/8 yep there's other ways of doing it also but like if you could draw it going baby steps it's the best option even though you sometimes like have to be very deliberate okay so that's enough of that that's the joint probability distribution

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Keywords: joint probability distribution, marginal distribution, joint distribution, marginal probability, marginal probability distribution, joint probability, joint distributions, joint and marginal probability, joint and marginal distributions, bivariate probability distribution, joint probability distribution example, marginal pmf, bivariate distribution, joint probability function, marginal distribution function

Id: Om68Hkd7pfw

Length: 13min 12sec (792 seconds)

Published: Sat Nov 14 2015