0
$\begingroup$

I'm stumped on the following problem:

"Use the conditional variance formula to determine the variance of a geometric random variable $X$ having parameter $p$."

My original idea was to have a variable $Y$ where $Y = 1$ if the first trial is a success and $Y = 0$ if not, and then to condition $X$ on $Y$:

$$\mathrm{Var}(X) = E\big[ \mathrm{Var}(X|Y) \big] + \mathrm{Var}\big( E[X|Y] \big)$$

So I begin to calculate: $$ \mathrm{Var}(X|Y = y) = E\big[ (X - E[X|Y = y])^2 | Y = y \big]$$

At which point I'm kind of lost, as I'm not sure how to calculate the conditional probabilities such as E[X | Y=y]. Am I heading in the right direction? If so, how do I calculate these conditional probabilities?

$\endgroup$
2
$\begingroup$

You are heading in the right direction. If $Y$ is as you've defined, then $E(X\mid Y)$ and $\operatorname{Var}(X\mid Y)$ are both variables that take two possible values. To determine these values, think about the two cases:

  • $Y=1$, which occurs with probability $p$. Then the first toss is a success so conditional on $Y=1$ we know that $X=1$, i.e. we hit the success on the first trial. Therefore $E(X \mid Y=1)=1$ and $\operatorname{Var}(X\mid Y=1)=0$.

  • $Y=0$, which occurs with probability $1-p$. Then the first toss was a failure, so we need to keep tossing. We can write $X=1+X'$ where $X'$, the remaining tosses required to see success, has the same distribution as $X$ but is independent of the first toss. Therefore $E(X\mid Y=0)=1+E(X)=1+\frac1p$ and $\operatorname{Var}(X\mid Y=0)=\operatorname{Var}(X)$.

Now apply the following fact, which you can easily verify:

Fact: If $W$ is a variable taking value $a$ with probability $p$ and $b$ with probability $1-p$, then $E(W)=(a-b)p$ and $\operatorname{Var}(W)=(a-b)^2p(1-p)$.

Using this fact, and plugging into the conditional variance formula, we obtain:

$$\operatorname{Var}(X)= (1-p)\operatorname{Var}(X) + (\textstyle\frac1p)^2p(1-p).$$

Now solve for $\operatorname{Var}(X)$.

$\endgroup$
1
$\begingroup$

The key is that if the first trial fails you are at the same place you began plus one failure. However, if the trial succeeds you are done.

$$\begin{align}\mathsf E(\mathsf {Var}(X\mid Y)) ~=~& \mathsf E(\mathsf E(X^2\mid Y)-\mathsf E(X\mid Y)^2)) \\ =~ & \mathsf P(Y{=}0)(\mathsf E(X^2\mid Y{=}0)-\mathsf E(X\mid Y{=}0)^2)+\mathsf P(Y{=}1)(\mathsf E(X^2\mid Y{=}1)-\mathsf E(X\mid Y{=}1)^2) \\ =~ & \mathsf P(Y{=}0)\big(\mathsf E((X+1)^2)-\mathsf E(X+1)^2\big)+\mathsf P(Y{=}1)(0) \\ =~ & (1-p)\big(\mathsf E(X^2)+2\mathsf E(X)+1-\mathsf E(X)^2-2\mathsf E(X)-1)\big) \\ =~ & (1-p)\big(\mathsf E(X^2)-\mathsf E(X)^2)\big) \\ =~ & (1-p)\mathsf {Var}(X) \end{align}$$

Can you continue?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.