So last time we were talking
about standard normal, right? Normal zero one. So just a few quick facts
that we proved last time. So our notation is,
traditionally it's often called Z, but I'm not saying Z has
to be standard normal. Or you have to call standard normal Z,
just we often use letter Z for that. If Z is standard normal, then first
of all, we found its PDF, right? We figured out the normalizing
constant,it's CDF. It's CDF you can't actually
do in closed form, so therefore it's just called capital Phi. That's just the standard notation for
the CDF. We computed the mean and
the variance last time. Remember, the mean E of Z = 0. That's just immediate by symmetry. Then we also did the variance. Variance equals in this case
the variance is E of Z squared equals 1. Cuz variance of E of Z squared
minus E of Z squared the other way, but that's 0, so that's one. That we computed last time
using integration by parts, so we did that last time. And if we wanted, this is by the way it's
called the first moment, second moment. If we wanted E of Z cubed,
this we didn't talk about last time. That's gonna be 0 again. Because, I'll just write
down what it would be. By LOTUS, we would have the integral minus
infinity, infinity 1 over root 2 pi, E to the minus Z squared over 2 dz. This integrates 1,
that would be just integrating the PDF. And LOTUS says if we want E of Z cubed,
we just stick in a Z cubed here. If we just wanted to do E of Z,
we'd put Z, if we want E of Z cubed, we'd put Z cubed, that's LOTUS. But this is just equal to 0,
because this is an odd function, again. So we talked about that in this case, but
the same argument would apply here for Z cubed. Similarly, for any odd power here, 5,
7, and so on, we'll immediately get 0. So this is called the third moment. At some point later in the semester we
can talk about where does the word moment come from. But that's just that's just terminology
for that that's called the third moment. E of Z cubed that would be called the
second moment first moment then and so on. Okay so in other words, by symmetry we already know that all
the odd moments of the normal are 0. The even moments well, we have this the second one if
we wanted E of Z to the fourth. Well it's going to be the integral except
put Z to the fourth instead of Z cubed, then that's not such an easy
integral anymore,okay? And it's not an integral that you need
to know how to do it at this point, we'll probably come back to how
to do things like that later, not before the midterm though. But at least you should immediately
know LOTUS that you could write down the integral for
E of Z to the fourth, it's just that happens to be an integral that I don't
expect that anyone could do right now. But at least you could write
down the integral, okay? Odd moments though, you just immediately
get 0 by symmetry, no integrals needed. Okay so I was talking about symmetry,
let me just mention symmetry one other way which is that minus Z is
also standard normal. And that's just another way to
express the symmetry of it. That is, the PDF is this bell
curve that's symmetrical about 0. So if you flip,
this flips between plus and minus, right. Just flipping the sign,
that changes the random variable, it makes a positive into negative,
makes negative into positive. But it does not change the distribution,
that's what the symmetry says. So you can either just see this by
symmetry or you could compute the PDF of this by first find the CDF, then find
the PDF, and you'll see that that's true. That's a very useful fact.,
it's always useful looking for symmetries. Okay, so this is just stuff
about the standard normal. But now we wanna introduce
what happens with normal where this is not necessarily 0, 1, okay? So this is the general normal. We say that X,
if we let X equal mu plus sigma Z where mu is any real number and we would call that the mean cuz
that's going to be the mean. But we would also call that the location. Because we're just adding a constant,
it means a shift in location. We're not changing what the density
looks like by adding mu, we're just moving it around left and
right. And sigma is any positive number,
mu could be negative, sigma has to be positive, and
that's called the standard deviation. Remember standard deviation we defined
as the square root of variance. So sigma is the standard deviation but
we also call that the scale because we're just rescaling everything
by multiplying by a constant. So that's gonna effect if
you draw one of the density, it's gonna effect how wide or
how narrow that curve is. It still has to integrate to 1, so you
can't just make it really big and wide and suddenly you made the area blow up. You also have to make sure that you
multiply by a normalizing constant so it still integrates to 1, but you can
still make it more wide or more narrow. Okay then we say Then we say X is normal with mean mu and variance sigma squared. So those are the two parameters. So the reason most books would do
this a little bit differently and start by writing down the PDF of this. But this is a more useful and
more insightful way to think about it, where we're saying there's just one
fundamental basic normal distribution. That's what we call the standard normal. Once we understand the standard normal
we can easily get any other normal distribution we want just by multiplying
by a constant adding a constant. So it's reducing everything back
down to the standard normal. That's really useful to always
keep that in mind instead of just looking at ugly formulas, okay? So let's actually check that this
has the desired mean and variance. So obviously the expected value of
X just by linearity with mu plus sigma expected value of Z is 0, so
that's just mu, just immediate from this. For the variance, Then we need to talk a little bit more about what happens,
what are the properties of variance. So I'll come back to this in a minute. First, let's just talk a little bit
more in general about variance. We did a quick introduction
to variance before but we should go a little bit further. So remember,
there's two ways to write variance. The definition is to subtract
off the mean, square it, the average distance
squared of X from its mean. But we also showed that can also be
written as E(X) squared, this way, minus E(X) squared the other way, okay? Now in particular, if we had the variance
of X plus a constant, intuitively, if we just add a constant we're not
changing how variable X is, right? So intuitively that should be
the same as the variance of X. And you can see that immediately from this
first formula because You replace by x by x + c, and the mean also shifts by c by
linearity, you get the exact same thing. So that's immediate from this, so adding
a constant has no effect on the variance. Now if we multiply by a constant,
then from either of these formulas, just imagine sticking in a c here and
a c here. But the c comes out because of linearity
again, but it's squared, then. So the variance of a c times x is
c squared times the variance of x. And a common mistake is to
forget the square here, but that really messes things up, so
variance is coming out with the square. And an easy way to see that is,
if c is negative, this is still valid. But if you forgot to write the square
here, you would get a negative variance. If you ever get a negative variance,
that's very, very bad, variance cannot be negative. So anytime you compute a variance,
the first thing you should check is, is the thing I wrote down
at least non-negative? And the only case where it could
be 0 is if it's a constant, so it's always greater than or equal to 0. And variance of X = 0 if and only if X is a constant
with probability 1. P of X = a = 1 for some a, that is, with probability 0,
something bad could happen. But with probability 1,
it always equals this constant a. So that would have variance
0 because the stuff with probability 0 doesn't affect
anything, so essentially it's a constant. If it's not a constant,
the variance will be strictly positive. Okay, so that's the variance of a constant
times x, and then just one other factor about, we'll do a lot more with
variance like after the midterm. But only one other thing to point out for
now is that variance, unlike expected values,
variance is not linear. So variance of x + y is not equal
to variance x plus variance of y. In general, it may be equal, but
it's not necessarily equal, so actually, it violates both
of the linearity properties. If it were linear, we would want
constants to come out as themselves, and here it comes out squared. And we can't say the variance of
the sum is the sum of the variances. It is equal,
we're not gonna show this until later, we'll show this at sometime
after the midterm. It is equal if x and
y are independent, but remember, linearity holds regardless of whether the
random variables are independent or not. So if they're independent, it will
be equal, we'll show that later, but in general, they're not equal. And one quick example of that would be,
what if we look at the variance of x + x? All right, that's an extreme case
of dependence, that's when x, it's actually the same thing, right? Well, the variance of x +
x Is the variance of 2x, which we just said is 4
times the variance of x. So if this were true, if this were equal,
we would get 2 times the variance of x. And this says we get 4 times the
variability, not 2 times the variability, but that's just a simple example of that. But that's also a common mistake
that I've seen before when students are dealing with, in the past I've
asked questions either on homeworks or exams where we have something like 2x. And a lot of students took
the approach of, well, 2x is x + x. Of course, that's valid,
but then at that point, they made the mistake of replacing
x + x by, let's say, x1 + x2. Where those are IID,
with the same distribution as x. That's completely wrong because
x is not IID with itself. It's extremely dependent and then somehow
replacing it by independent copies, then it doesn't work. So I'm telling you to be careful of this, just keeping track of
dependents versus independents. Here they're extremely dependent, and
so that's why we got this 4 here. And I think, intuitively,
that should make some sense, right? If this was like x1 and x2 and they're independent,
then the variabilities just add. Here, they're exactly the same, so that magnifies the variability, okay. So that's a few quick
notes about variance, so now coming back to this for
the normal case. We just saw that adding mu does nothing
to the variance, multiplying by sigma. Then it comes out as sigma squared, that's
sigma squared times the variance of z. Well, that's just sigma squared,
okay, so that confirms that when we write this, this is the mean and
this is the variance. So those are the two parameters
of the normal distribution. Ane whenever you have
a normal distribution, you should always think about
reducing it back to standard normal. So we could also go the other way around,
and I don't need much space for this. Because this is just, I'm just
gonna solve this equation for z, so if we do it the other way, solve for z. z equals x minus mu over sigma,
very easy algebra, that's called standardization. So standardization says,
I'm just going the other direction here. I was starting with the standard normal,
and we can construct a general
normal this way. Now what if we wanted to go the other way,
we started with x, which is normal mu sigma squared. Subtract the mean divided
by the standard deviation, and that will always give
us a standard normal. So that process is called standardization,
it's very, very useful, it's simple, right, just subtract the mean
divided by the standard deviation. And yet sometimes students
get confused about it, or divide by the variance instead of
dividing by the standard deviation, or just don't think to do
it in the first place. So that's why I'm emphasizing that,
it's a simple but useful transformation. Okay, so
as a quick example of how we use that, let's derive the PDF
of the general normal. Find PDF of normal mu sigma squared, well, one way to find it is to
look it up in a book. But that doesn't tell you anything,
that's just like a formula in a book. So what we want to understand is,
assuming that we already know the PDF of the standard normal, how can we get
the PDF of the non-standard normal? In a way, that's easy,
without having to memorize stuff, okay, so let's call this x again,
so let's find the CDF first. So by definition,
this is just good practice with CDFs. Everyone here should make sure that
you're good at CDFs and PDFs and PMFs. And that just takes practice, so this
is just some simple practice with that. By definition, the CDF is this, and now I just told you that a useful trick is
to standardize, so let's standardize this. It's the same thing as saying X
minus mu over sigma is less than or equal to lowercase x minus mu over sigma,
right. Sigma is positive, so it doesn't
flip the inequality to do that, so I standardized it. The reason I standardized
it was because now, this thing on the left is standard normal. So by definition, this is just the CDF
of the standard normal evaluated here. So by definition,
we immediately know that's just capital phi of x minus mu over sigma,
now to get the PDF, To get the PDF we just have to
take the derivative of the CDF. That's just the chain rule right, because
this capital phi is the outer function and then we have this inner function here so it's just the chain rule
from basic calculus. It's the derivative of the outer
function evaluated here, times the derivative
of the inner function. The derivative of this inner function
is just 1 over sigma, right, cuz 1 over sigma times x. So we are gonna get a 1 over sigma, and
then we are gonna get the derivative of this the derivative of capital phi is
just the standard normal PDF, right? And it says evaluated here, so I'm just
gonna write down the standard normal PDF, and I'm gonna evaluate
it at x- mu over sigma. And that's it, we're done. So it should be a very, very quick calculation in
order to be able to do that. And as another quick example. Let's say over here in the corner,
we said what happens, z is standard normal, what happens to -z? Let's also ask the question
of what happens to -x? Well, you could work through
a similar calculation, but I think the neatest way to think of it is,
we're thinking of x as mu + sigma z. So -x- mu + sigma times -z. But -z is standard normal. So this is just of the form some location constant plus sigma
times the standard normal. So we immediately know that's
normal -mu sigma squared. Which again, makes sense intuitively,
because we put a minus sign, so we put a minus sign on the mean. We do not put a minus
sign on the variance, because variance can't be negative,
so the variants stay sigma squared. So you could do a calculation for
this, but this is just immediate from thinking
of x in terms of the standard normal. So this is the easiest way to do this,
okay? And a useful fact just to know, but
we'll prove this much later in the course. Later we'll show that if x1, let's say xj is normal mu j, sigma j squared, and they're independent. Let's say for j equals 1 to 2. Then, x1 + x2 is normal, mu1 + mu2 sigma 1 squared + sigma 2 squared. So that's something we need to prove,
and we'll do that much later. The sum of independent normals is normal,
but the reason I'm mentioning it now is just let's think about what
happens to the mean and variance. By linearity, we know that the mean
would have to be mu1 + mu2. Variance, this is something
else we'll prove later. In the independent case we can
just add up the variances, so it's juts sigma1 squared + sigma2 squared. Now what if we looked at x1- x2? I'm mentioning this now, because I can't even count the number of
times when I've seen the mean is mu1- mu2. That's just linearity again. I can't even count the number
of times I've seen students write that the variance
is sigma1 squared- sigma2 squared. Well, first of all that could be negative,
so that doesn't make any sense. And secondly, any time you see
a subtraction you can really think of that as adding the negative of something,
right? So this is + of -x2. And -x2 still has variance sigma2 squared,
so the variance is still add. That's just a useful fact to keep in mind,
we'll prove it later. But I'm mainly talking about right
now just in terms of what happens to the mean and variance. Later we'll see why are they still normal. That's just one very useful
property of the normal. So let's just do a lot of things without
leaving the realm of normality, right? If you added two of them and then it
somehow becomes some completely different distribution, it's gonna
be hard to work with. So that's a very nice
property of the normal. Okay, one other fact about the normal that's just like a rule of thumb for
the normal. Because of the fact that you can't
actually compute this function, capital phi other than by
having a table of values. Or a computer that, or a calculator that specifically
knows how to do that function. You can't do it in terms
of other functions, it's useful to just have a few
quick rules of thumb, so there's something called
the 68- 95-99.7% rule. And I don't know who named it that,
but at the first time I heard of this that's the stupidest name for
a rule that I have ever heard of. However, then I always remember that,
so actually it works very well. It simply says it's just the three
simple numbers telling us how likely is it that a normal
random variable will be a certain distance from its mean measured
in terms of standard deviation. So this says that, if x is normal, then the statement is that the probability that x is more than 1 standard
deviation from its mean. So notationally we would
just write it like that. But intuitively, that's just saying what's
the chance that it falls more than 1 standard deviation, right? That's 1 standard deviation. This would say the distance is more than
1 standard deviation away from the mean. Well, I was actually right the other way. The probability that x is
within 1 standard deviation of its mean is about 68%. The chance that x is within 2 standard
deviations of its mean is about 95%. And the chance that it's same with 3
standard deviations is about 99.7%. So, in other words, it's very common for people in practice to add and
subtract 2 standard deviations. What that's saying is for the normal,
that's gonna have 95% chance of so, let's say you got a bunch of observations
from this distribution independently. We would expect about 95% of them
are gonna be within 2 standard deviations of the mean, 99.7% within 3. So you can convert these statements into
statements about capital phi which is good practice while just making sure you
understand what capital phi is. But basically, this is just a few values
of capital phi just written in kind of a more intuitive way. Okay, so that's all for
the normal distribution. So the main thing left to
talk more about is LOTUS, and a couple examples of LOTUS and
using LOTUS to compute variances. For example, we proved that the variance of
the Poisson is Poisson lambda has, sorry. We proved that the mean of
a Poisson lambda is lambda. We have not yet
derived the variance of a Poisson lambda. So that's definitely
something we should do. So, okay. So let's do the variance of the Poisson. And that will also give us
a change to understand more, what's really going on with LOTUS? Why does LOTUS really work? So suppose we had a random
variable such as the Poisson, but right now I'm just thinking in general. A random variable who's possible values
are zero, one, two, three, and so on. So let's call our random variable x. And x can be 0, 1, 2, 3, etc, okay? And suppose that its pmf. To say what the pmf is I just need
to say what's the probability of, 0 let's call that P0 probability of 1,
P1, P2, P3. So all I did here was write out the pmf, just stringing it out as a sequence,
right? But that's just specifying the pmf and I'm calling them pj is
the probability that x equals j. Now to figure out variance we
need to study x-squared, right? So let's look at x squared. So 0-squared is 0,
1-squared is 1, 2-squared is 4, 3-squared is 9, and
we keep going like that. From this point of view, it should
be easy to see what we should do. Because E(x), remember for a discrete random variable E(x)
is the sum of x times the pmf. Now here we want E(x-squared), but notice that the probability
that x-squared equals say 3-squared is just the probability P3 of
being in this column here, right? So the probabilities didn't change, and we could just still use x-squared times
the probability that x = x, right? Because when an x-squared
takes on these possible values with these same probabilities. That's what LOTUS is saying, so
it's pretty intuitive in that sense. The case that you have to think more about
is the case where this function is not 1 to 1. So now squaring is not 1 to 1 in general. If I had had negative numbers,
then you would have duplicates here and you would have to sort that out. What LOTUS says is even when you have
those duplications, this still works. That I think is a little less obvious, if you think about it you can see why it's
true, but it's not completely obvious. In this case,
because we're not non-negative anyway, this is one to one and
then it just immediately true, okay? But LOTUS this is saying, no matter
how complicated your function is, something kind of this flavor still works, regardless of whether
you have duplications. So now we're ready to get
the Poisson variance. So this is just in general if you have a random variable
non-negative integer values. Now let's look at the specific
case of Poisson lambda and we want to find E(x) squared. And according to LOTUS we can just
write that as the sum k = 0 to infinity k-squared E to the minus lambda, lambda to the k over k factorial,
that's the pmf. So we have to figure
out how to do this sum, and this looks like
a pretty unfamiliar sum. I mean my first thought when I see this
would be, well this is k times k and we can cancel and
get a k minus one factorial here. And there's nothing wrong
with doing that but it's still kind of annoying because
we still have k-squared up here. When we were just planning the mean,
then we just had a k and we cancelled it and things are nice. But now we have a k-squared,
it's more annoying, okay? So here's another method for
dealing with something like that. The general method is start
with what we know, right? So what we know how to do is
the Taylor series for e to the x. Hopefully you all know that by now,
we keep using it over and over again. The sum, I'll write it in terms of lambda. The sum of lambda to
the k over k factorial. Is e to the lambda, and this is valid for
all real lambda, even for imaginary numbers, complex numbers,
this is always true, always converges. Now if I wanna get a k in front, then a natural strategy would be to
take the derivative of both sides. Well that's pretty nice right, because the derivative e to
the lambda is e to the lambda. The derivative of the left-hand side, I'll start the sum at 1
now because at 0 it's 0. So we have k lambda to
the k -1 over k factorial. I just took the derivative of both sides. I exchanged the derivative and the sum, which is valid under some
mild technical conditions. Now we're getting closer, but we still
only have a k, not a k squared, okay? So my first impulse would be,
take a derivative again, that's slightly annoying cuz then I'd get
a k-1 coming down, I want a k, not a k-1. So to fix that, all we have to do is
multiply both sides by lambda, okay? So, just put lambda on both sides. So I call that replenishing the lambdas. We just replenish it,
that we have a lambda there. I'll write it again,
k equals one to infinity. K, lambda to the k over k factorial
equals lambda e to the lambda. We've replenished our supply of lambda's, now we can take the derivative again and
we have what we want. Okay, so
I take the derivative a second time and k = 1 to infinity, take the derivative
again, now it's k-squared. Lambda to the k- 1 over k factorial. Well now we have to use the product rule,
the derivative of lambda, e to the lambda is lambda e to the lambda plus e
to the lambda by the product rule. Which we can factor out as e to
the lambda times lambda + 1. Okay, well that's exactly
the sum that we needed. Cuz this e to the minus lambda comes out,
so this is e to the minus lambda,
e to the lambda, lambda + 1. I'm missing some,
is there another lambda somewhere? Lets see, we have to replenish it again. Just put a lambda. Okay, so here we have lambda to the k- 1,
there we want lambda to the k. So its replenish again,
then there is another lambda there, okay. I'm just bringing this k- 1 back
to being lambda to the k, right? So that's just lambda squared + lambda. And now we have the variance. So the variance of X equals this thing, lambda squared plus lambda
minus the square of the mean, which is lambda squared equals lambda. So this course is not really
about memorizing formulas, but that's one that's very easy and
useful to remember. The Poisson lambda has mean lambda,
and has variance lambda. So that's kind of a strange
property if you think about it. That the mean equals the variance,
it's a little bit, maybe it would seem more natural if
the mean equal the standard deviation or something like that, because then
those are kind of in the same scale. But Poisson,
it doesn't actually have units. Poisson is just counting
numbers of things, so it doesn't have that some
dimensional interpretation. So, yeah, I wanted to also mention
that about standardization as well. Another reason this thing is really
nice to work with in the normal is if you think of normal as being
a continuous measurement in some unit, it could be a unit of length,
time, mass, whatever. If x is measured in
whatever unit you want, let's say it's time measured in seconds,
then that's seconds minus seconds divided by seconds,
the seconds cancel out. That means this is a dimensionless
quantity, which is part of what's making this standardization, it's kind of
making it more directly interpretable instead of having to worry about whether
you measured it in seconds or years. So if we started with one measurement in
seconds and one measurement in years and standardized both of them,
we get the same thing. The same measurement in different units. So that's a nice property of that. Okay, so
that's the variance of the Poisson. We haven't yet gotten the variance of
the binomial, so I'd like to do that. There's an easy way and a hard way. Well, except the hard way I don't think,
actually sorry, there's three ways to do it. There's a really easy
way that we can't do yet because we haven't proven
the necessary fact. There's an easy way that we can do,
so that's what I'm gonna do. And then there's an annoying way,
which we're not gonna do. The annoying but direct is we
want the variance of a binomial. We wanna find the variance. The most direct obvious way to do this
would be to use lotus to get E(x squared) which would mean you would have to
write down something like this, except here we wrote the Poisson PMF. Instead you'd have to write n choose k,
p to the k, whatever, the binomial PMF, right. And then you'd have to do that sum. And you can do it, but
that's pretty tedious. And you have to figure out how to do
that sum and do a lot of algebra. Okay, so
that's the way I don't wanna do it. The easiest way to do it would
be using this fact here. Which is that the variance of a sum
of independent things is the sum of the variance,
if they're independent, right. That's if, okay. So the easiest one,
we haven't proven this yet so this is not valid to do it at this way
right now but just kinda foreshadowing. We can think of the binomial, we've emphasized the fact that we can
think of a binomial as the sum of n independent Bernoulli p. So once we prove this fact,
that's applicable. So all we have to do is get
the variance of Bernoulli p, which is a really easy calculation
cuz the Bernoulli is just zero one, so that's a very very easy calculation. To get the variance of a Bernoulli p and
multiply by n, that's the neatest way to do it. You can do it that way in your head
once we get to that point, okay. Now here's kind of the compromise method
which is also just good practice with other concepts we've done,
especially indicator random variables. So I'm still going to use the same idea of representing x as a sum
of Iid Bernoulli p. So I'll write them as I1 plus blah,
blah, blah, plus In, just to emphasize the fact that
they're indicators, I for indicator. Where Ijs are Iid Bournulli p, right. So we've been doing this
many times already. That's just an indicator of success
on the jth trial, add up those and we get a binomial. Okay, so
now if we want the expected value of x squared, Let's just square this thing. Let's actually not do
the expected value yet. We'll just square it then
take the expected value. So just square this thing. Well you know you do i1 squared and
just square all the things, right. So it's i1 squared plus blah
blah blah plus In squared plus, but as you know we get a lot
of cross terms, right. Your imagining this big thing times
itself, so every possible cross term, each one twice, you have 2I 1I 2 and
2I 1I 3 and so on. All possible cross terms and
each cross term has 2 in front. Just like when you square x+y,
you get x squared + y squared + 2xy. We get all these cross terms. It doesn't matter what
order we write them in. Maybe we've ordered them in this way. So that's the last one. It doesn't matter the order. Okay, so it's all the cross terms. That looks pretty complicated. But it's actually much
simpler than it looks. Now let's take the expected value
of both sides, use linearity. Of the same,
this is a good review example as well. We're using the same tricks, symmetry,
indicator, random variables, and linearity. Each of these, these are Iid. So by symmetry,
this is just n times anyone of them. So let's just say nE(1 squared). That's just immediate by symmetry, right. So we don't have to write that big sum,
just n times one of them. And how let's just count
how many of these, well there's n choose two cross terms,
right. Because for any pair of
subscripts we have a cross term. So it's really just 2(n choose 2),
and then just take one of them for concreteness, let's say E(I1I2). Now this is even nicer,
well it definitely is looking better. But this is even better than it looks
because I1 is either just 1 or 0. If you square one you got one,
if you square zero you got zero. So I1 squared is just I1. So E(I1), that's just
the expectorate of Bernoulli p is p. So that's just np+n choose 2 is n times n-1 over 2,
so the 2s cancel. So this is really just n(n-1). Now let's think about this
indicator random variable. Well I called it an indicator
random variable, well actually it's a product
of indicator random variables. But actually a product of indicator random
variables is an indicator random variable. This thing here is the indicator
of success on both the first and the second trial, right. Because if you think of multiplying
two numbers that are zero and one, you get zero if at least one of these is zero,
you would get one if they're both one. So that's the indicator
of success on both. So it's a product but
it's actually just one indicator. Success on both trials, number 1 and 2. So its expected value is just
the probability of that happening. That probability of success
on both the first trial and the second trial, because the trials
are independent, is just p squared. Okay, so that's just, so what we just computed is
the second moment of the binomial. That's np+, if we multiply that np+n squared p squared-np squared, right. Now to get the variance all we have to do
is subtract, The square of the mean, okay. So we showed before that
a binomial np has mean n times p. So if we square that, that's this term
n squared p squared, so that cancels. So we're just canceling
out this middle term and we just have np- np squared = np (1- p). Which we would often write as npq with q = 1- p. So binomial variance is npq. So that's just a good review of indicator
of random variables and all of that stuff. So now we know the variance
of the Poisson, the normal, the uniform, the binomial. For the geometric, it's kind of a similar calculation, we did the mean
of the geometric in two different ways. The flavor of the calculation is similar
to this except we have a geometric series instead of the Taylor series for
e to the x. So I don't think it's
worth doing that in class. So in general hypergeometric, let's
talk a little bit about hypergeometric, that's pretty nasty. In the sense that in the hypergeometric, we could write it as a sum of
indicator random variables. We're imagining we're drawing
balls one at a time and, or picking elk one at a time and
success is getting a tagged elk. But the problem is that
they're not independent. So as far as the mean is
concerned we still use linearity. For the variance it's more complicated. So we'll worry about the variance of
a hypergeometric after the midterm. That's more complicated. But for the binomial this is really,
well, actually we could still. Here I didn't actually use the factor
there independent cuz I was just using linearity. So you could use a similar approach, so
actually you could do it this way, but it would be too tedious to do it
like on a midterm or something. But you could square it, if these
are dependent well, you can still work out the probability that the first two
elk that you pick are both tagged. You could do that without
too much trouble. But it's pretty messy looking. All right, so that's variance,
and I guess the last thing to do is just to explain more
about why is LOTUS true? And the basic proof of
that is actually kind of conceptually similar to
how we proved linearity. So we're trying to prove LOTUS, and
I'm only gonna prove it for a discrete. Let's say discrete sample space. That's the case where I'm
imagining finitely many pebbles. In the general case the ideas
are not essentially different. It's just that we kind of need to
write down some fancier integrals and use more kind of more technical math,
but the concept is similar. So this is enough to give you the idea. So for discrete sample space,
so the statement is that the expected value,
that's all we are trying to show, is that the E(g(x)) can be written
as the sum of g(x) P(X=x). So right, we can use the PMF of
x we do not have to first work on figuring out the distribution of g(x). That's all we are trying to do,
so let's think about it. Let's think about it as a sum of,
sum over the other. We could sum the other way around, sorry. Let me say this a different way, let me remind you of the identity
that we use for proving linearity. That was this group versus ungroup thing. So what we have is two different
ways to write a certain sum. We could either write this thing,
g(x)P(X=x) or we could write it the other way,
which is a sum over all s. Each s, we're thinking of that
as s in the sample space S. So each little s is a pebble. And if we're summing it
up pebble by pebble, then what we're doing is remember
random variables are functions. So, and g(x) just means we apply
the function x then apply the function g. So we're just computing g(x(s)), that's just the definition
times the mass of that pebble. So. If you stare at this equation long enough,
and we have five minutes left to stare at
that equation, so that's plenty of time. This is why LOTUS is true. It's just a matter of
understanding this equation. So I'm gonna talk a little more about,
how do you make sense of this equation? This is the grouped case. This is the ungrouped case. Remember I talked about pebbles and
super pebbles, ungrouped. This says take each pebble,
compute this function, g of x of s, and
you take a weighted average. Those are the weights. This says,
first combine all of the pebbles that have the same value of x into you know,
super-pebbles. A super-pebble means we grouped together
all pebbles with the same x value, not the same g(x) value, the same x value. Group those together then average,
you get the same thing. So if I want to write that out
in a little bit more detail. One way to think of it is as a double sum,
right? Because we could imagine
first summing over x. I'm gonna break this sum up. What I just explained to you was the
intuition for why this is equal to this. Because we're just grouping them
together in different ways so we changed the weights around, but as long
as we changed the weights appropriately we should get the same average. That's the intuition. But for any of you who wanna see more of
an algebraic reason, justification for that, the way to think of
it is as a double sum. So the double sum would be, I mean to rewrite this says
sum overall pebbles right? But one way to think of that would be
first sum over values of little x. And then for each value of little x,
sum over all pebbles, s such that x(s) = x. Because this is just a sum
of a bunch of a numbers. We can sum them in any order we want. So I can rearrange them, in this particular order where I'm saying
first sum over the little x values, and then group together, and sum over all
the pebbles that have that value. It's the exact same thing,
I just reordered the terms. So that's g(x(s)) times P(s). Now let's just simplify this double sum. The reason I wanted to write it as
a double sum like this is that within this inner summation X(s)=, so
this thing is just g(x). The cool thing is that g(x) does
not depend on s so that comes out. So we actually have the sum
over x of g(x) times the sum of what ever is left p(s). And now so that's summed over
all s such that s(x) = x. And now we're done with
the proof because this sum here is just saying add up
the masses of all the pebbles labeled x. In other words,
that's what I called a super pebble. The super pebble,
the mass is the sum of all the masses of the little pebbles
that form the super pebble. That's p of, this is just practice, this is going back to the very beginning
of, events and what's a random variable. That's just the event X = x. We talked, what does it mean for
big X to equal little x, right? What does that equation mean? That's an event. That's this event that we have here. Okay, so that's why that's true. So that's why LOTUS is true. Anyway that's all for
now and Friday we'll review, let me know if you have any suggestions
for things to do on Friday.
