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In this segment, we will derive
the formula for the

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variance of the geometric PMF.

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The argument will be very much
similar to the argument that

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we used to drive the expected
value of the geometric PMF.

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And it relies on the
memorylessness properties of

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geometric random variables.

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So let X be a geometric
random variable with

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some parameter p.

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The way to think about X is like
the number of coin flips

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that it takes until we obtain
heads for the first time,

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where p is the probability
of heads at each toss.

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Recall now the memorylessness
property.

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If I tell you that X
is bigger than 1--

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which means that the first
trial was a failure---

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we obtained tails.

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Given that event, the remaining
number of tosses has

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the same geometric PMF
as if we were just

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starting at this time.

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So it has the same geometric PMF
as the unconditional PMF

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of X. And this is the property
that we exploited in order to

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find the expected value of X.

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Now let us take this observation
and add one to the

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random variables involved and
turn this statement to the

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following version.

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The conditional PMF of X--

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which is this random
variable plus 1--

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is the same as the unconditional
PMF of this

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random variable plus 1.

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So it's the same statement as
before except that we added 1.

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One consequence of the
memorylessness that we have

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already seen and exploited is
that the expected value of X

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in the conditional universe
where the first coin flip was

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wasted is equal to 1--

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that's the wasted coin flip--

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plus how long you expect to have
to flip the coin until

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you obtain heads for the
first time, starting

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from the second flip.

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Since the conditional
distribution of X in this

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universe is the same as the
unconditional distribution of

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this random variable, it means
that the corresponding

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expected value in this universe
is going to be equal

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to the expected value of this
random variable, which is 1

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plus the expected value of X.
And by exactly the same

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argument, the random variable
X squared has the same

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distribution in the conditional
universe as the

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random variable X plus
1 squared in the

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unconditional universe.

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So since X in the conditional
universe has the same

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distribution as X plus 1, it
means that X squared in the

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conditional universe has the
same distribution as X plus 1

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squared in the unconditional
universe.

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So now let us take those facts
and use a divide and conquer

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method to calculate the expected
value of X squared.

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We will use exactly the same
method that we used in order

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to calculate the
expected value.

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We separate into
two scenarios.

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In one scenario, X
is equal to 1.

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And then we have the expected
value of X squared given that

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X is equal to 1.

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And then we have another
scenario--

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the scenario that X
is bigger than 1.

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And then we have the expected
value of X squared given that

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X is bigger than 1.

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So this is just the total
expectation theorem.

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Now let us calculate terms.

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The probability that the first
toss results in success, that

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X is equal to 1-- this is p.

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And if X is equal to 1, then the
value of X squared is also

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equal to 1.

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And then there is probability 1
minus p that the first trial

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was not a success.

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So we get to continue.

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We have this conditional
expectation here.

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But it is equal to this
unconditional

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expectation up there.

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And now let us expand the terms
in this quadratic and

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write this as expected value
of X squared plus twice the

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expected value of X plus 1.

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Now we know what this expected
value here is.

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The expected value of a
geometric is just 1/p.

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And what we're left with is an
equation that involves a

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single unknown.

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Namely, this quantity
is the unknown.

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And we can solve this linear
equation for this unknown.

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We carry out some algebra,
which is not so

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interesting by itself.

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And after we carry out the
algebra, what we obtain is

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that the expected value of X
squared is equal to 2 over p

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squared minus 1 over p.

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And then we use the formula that
the variance of a random

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variable is equal to the
expected value of the square

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of that random variable
minus the square of

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the expected value.

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We already know what that
expected value is.

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We found the expected
value of the square.

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And putting all that together,
we obtain a final answer.

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And this is the expression for
the variance of a geometric

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random variable.

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It goes without saying that for
this calculation to make

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sense, we need to assume that
the parameter that we're

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dealing with is positive.