Hey, everyone.
Welcome back.
Today, we're going to do another fun problem that has
to do with a random number of coin flips.
So the experiment we're going to run is as follows.
We're given a fair six-sided die, and we roll it.
And then we take a fair coin, and we flip it the number of
times indicated by the die.
That is to say, if I roll a four on my die, then I flip
the coin four times.
And then we're interested in some statistics regarding the
number of heads that show up in our sequence.
In particular, we want to compute the expectation and
the variance of the number of heads that we see.
So the first step of this problem is to translate the
English to the math.
So we have to define some notation.
I went ahead and did that for us.
I defined n to be the outcome of the die role.
Now, since we flip the coin the number of times shown by
the die roll, n is equivalently the number of
flips that we perform.
And n, of course, is a random variable, and I've written its
PMF up here.
So Pn of n is just a discrete uniform random variable
between 1 and 6, because we're told that the die has six
sides and that it's fair.
Now, I also defined h to be the number of
heads that we see.
So that's the quantity of interest.
And it turns out that Bernoulli random variables
will be very helpful to us in this problem.
So I defined x sub i as 1 if the ith flip is
heads, and 0 otherwise.
And what we're going to do now is, we're going to use these x
sub i's to come up with an expression for h.
So if you want to count the number of heads, one possible
thing you could do is start with 0 and then look at the
first coin flip.
If it's heads, you add 1 to 0, which I'm going to call your
running sum.
If the first flip is tails, you add 0.
And similarly, after that, after every trial, if you see
heads, you add 1 to your running sum.
If you see a tails, you add 0.
And in that way, we can precisely compute h.
So the mathematical statement of what I just said is that h
is equal to x1 plus x2 plus x3, all the way
through x sub n.
So now, we are interested in computing e of h, the
expectation of h.
So your knee jerk reaction might be to say, oh, well, by
linearity of expectation, we know that this is an
expectation of x1, et cetera through the expectation of xn.
But in this case, you would actually be wrong.
Don't do that.
And the reason that this is not going to work for us is
because we're dealing with a random
number of random variables.
So each xi is a random variable.
And we have capital n of them.
But capital n is a random variable.
It denotes the outcome of our die roll.
So we actually cannot just take the sum of these
expectations.
Instead, we're going to have to condition on n and use
iterated expectation.
So this is the mathematical statement of what I just said.
And the reason why this works is because conditioning on n
will take us to the case that we already know how to deal
with, where we have a known number of random variables.
And of course, iterated expectations holds, as you saw
in lecture.
I will briefly mention here that the formula we're going
to derive is derived in the book.
And it was probably derived in lecture.
So if you want, you can just go to that formula
immediately.
But I think the derivation of the formula that we need is
quick and is helpful.
So I'm going to go through it quickly.
Let's do it over here.
Plugging in our running sum for h, we get this
expression--
x1 plus x2 et cetera plus xn, conditioned on n.
And this, of course, is n times the
expectation of x sub i.
So again, I'm going through this quickly,
because it's in the book.
But this step holds, because each of these xi's have the
same statistics.
They're all Bernoulli with parameter of 1/2, because our
coin is fair.
And so I used x sub i to say it doesn't really matter which
integer you pick for i, because the expectation of xi
is the same for all i.
So this now, the expectation of x sub i, this is just a
number, it's just some constant, so you can pull it
out of the expectation.
So you get the expectation of x sub i times the
expectation of n.
So I gave away the answer to this a second ago.
But x sub i is just a Bernoulli random variable with
parameter of success of 1/2.
And we know already that the expectation of such a random
variable is just p, or 1/2.
So this is 1/2 times expectation of n.
And now n we know is a discrete
uniform random variable.
And there's a formula that I'm going to use, which hopefully
some of you may remember.
If you have a discrete uniform random variable that takes on
values between a and b--
let's use w--
if you call this random variable w, then we have that
the variance of w is equal to b minus a times b minus a plus
2 divided by 12.
So that's the variance.
We don't actually need the variance, but we will need
this later.
And the expectation of w--
actually, let's just do it up here right
ahead for this problem.
Because we have a discrete uniform random variable, the
expectation is just the middle.
So you agree hopefully that the middle is right at 3.5,
which is also 7/2.
So this is times 7/2, which is equal to 7/4.
So we are done with part of part a.
I'm going to write this answer over here, so I can erase.
And we're going to do something very similar to
compute the variance.
To compute the variance, we are going to also
condition on n.
So we get rid of this source of randomness.
And then we're going to use law of total variance, which
you've also seen in lecture.
And again, the formula for this variance is
derived in the book.
So I'm going to go through it quickly.
But make sure you understand this derivation, because it
exercises a lot of stuff we taught you.
So this, just using law of total variance, is the
variance of expectation of h given n, plus the expectation
of the variance of h given n.
And now, plugging in this running sum
for h, you get this.
It's a mouthful to write.
Bear with me.
x1 through xn given n--
so I didn't do anything fancy.
I just plugged this into here.
So this term is similar to what we saw
in a previous problem.
By linearity of expectation and due to the fact that all
of the x i's are distributed in the same way, they have the
same expectation, this becomes n times the
expectation of x sub i.
And let's do this term over here.
This term--
well, conditioned on n, this n is known.
So we essentially have a finite known sum of
independent random variables.
We know that the variance of a sum of independent random
variables is the sum of the variances.
So this is the variance of x1 plus the variance of x2 et
cetera, plus the variance of xn.
And furthermore, again, because all of these xi's have
the same distribution, the variance is the same.
So we can actually write this as n times the variance of x
sub i, where x sub i just corresponds
to one of the trials.
It doesn't matter which one, because they all have the same
variance and expectation.
So now, we're almost home free.
This is just some scaler.
So we can take it out of the variance, but we
have to square it.
So this becomes expectation of xi squared times
the variance of n.
And then this variance is also just a scalar, so we can take
it outside.
So then we get variance of x sub i times expectation of n.
Now, we know that the expectation of x sub i is just
the probability of success, which is 1/2.
So we have 1/2 squared, or 1/4, times the variance of n.
So that's where this formula comes in handy.
b is equal to 6, a is equal to 1.
So we get that the variance of n is equal to 5 times--
and then 5 plus 2 is 7--
divided by 12.
So this is just a formula from the book that you guys
hopefully remember.
So we get 35/12.
And then the variance of xi, we know the variance of a
Bernoulli random variable is just p times 1 minus p.
So in our case, that's 1/2 times 1/2, which is 1/4.
So we get 1/4.
And then the expectation of n, we remember from our previous
computation, is just 7/2.
So I will let you guys do this arithmetic on your own time.
But the answer comes out to be 77/48.
So I will go ahead and put our answer over here--
77/48--
so that I can erase.
So I want you guys to start thinking about
part b while I erase.
Essentially, you do the same experiment that we did in part
a, except now we use two dice instead of one.
So in part b, just to repeat, you now have two dice.
You roll them.
You look at the outcome.
If you have an outcome of four on one die and six on another
die, then you flip the coin 10 times.
So it's the same exact experiment.
We're interested in the number of heads we want the
expectation and the variance.
But this step is now a little bit different.
Again, let's approach this by defining some notation first.
Now, I want to let n1 be the outcome of the first die.
And then you can let n2 be the outcome of the second die.
And we'll start with just that.
So one way you could approach this problem is say, OK, if n1
is the outcome of my first die and n2 is the outcome of my
second die, then the number of coin flips that I'm going to
make is n1 plus n2.
This is the total coin flips.
So you could just repeat the same exact math that we did in
part a, except everywhere that you see an n, you replace that
n with n1 plus n2.
So that will get you to your answer, but it will require
slightly more work.
We're going to think about this problem slightly
differently.
So the way we are thinking about it just now, we roll two
dice at the same time.
We add the results of the die rolls.
And then we flip the coin that number of times.
But another way you can think about this is, you roll one
die, and then you flip the coin the number of times shown
by that die and count the number of heads.
And then you take the second die and you roll it.
And then you flip the coin that many more times and count
the number of heads after that.
So you could define h1 to be number of heads in the first
n1 coin flips.
And you could just let h2 be the number of heads in the
last n2 coin flips.
So hopefully that terminology is not confusing you.
Essentially, what I'm saying is, n1 plus n2 means you'll
have n1 flips, followed by n2 flips, for a total
of n1 plus n2 flips.
And then within the first n1 flips, you can get some number
of heads, which we're calling h1.
And in the last n2 flips, you can get some number of heads,
which is h2.
So the total number of heads that we get at the end--
I'm going to call it h star--
is equal to h1 plus h2.
And what part b is really asking us for is the
expectation of h star and the variance of h star.
But here's where something really beautiful happens.
h1 and h2 are independent, and they are
statistically the same.
So the reason why they're independent is because--
well, first of all, all of our coin flips are independent.
And they're statistically the same, because the experiment
is exactly the same.
And everything's independent.
So instead of imagining one person rolling two die and
then summing the outcomes and flipping a coin that many
times and counting heads, you can imagine one person takes
one die and goes into one room.
A second person takes a second die and goes
into another room.
They run their experiments.
Then they report back to a third person
the number of heads.
And that person adds them together to get h star.
And in that scenario, everything is very clearly
independent.
So the expectation of h star--
you actually don't need independence for this part,
because linearly of expectation always holds.
But you get the expectation of h1 plus the expectation of h2.
And because these guys are statistically equivalent, this
is just two times the expectation of h.
And the expectation of h we calculated in part a.
So this is 2 times 7 over 4.
Now, for the variance, here's where the
independence comes in.
I'm actually going to write this somewhere where I don't
have to bend over.
So the variance of h star is equal to the variance of h1
plus the variance of h2 by independence.
And that's equal to 2 times the variance of h, because
they are statistically the same.
And the variance of h we computed already.
So this is just 2 times 77 over 48.
So the succient answer to part b is that both the mean and
the variance double from part A. So hopefully you guys
enjoyed this problem.
We covered a bunch of things.
So we saw how to deal with having a random number of
random variables.
Usually we have a fixed number of random variables.
In this problem, the number of random variables we were
adding together was itself random.
So to handle that, we conditioned on n.
And to compute expectation, we use iterated expectation.
To compute variance, we used law of total variance.
And then in part b, we were just a little bit clever.
We thought about how can we reinterpret this experiment to
reduce computation.
And we realized that part b is essentially two independent
trials of part a.
So both the mean and the variance should double.