0
$\begingroup$

I am stuck on the answer for this question: X is a geometric random variable, with var(X) = 6.

a) The conditional variance var(X-4 | X>4)? ans = 6.

b) Var(X-8 | X>4)? ans = 6.

I know that the variance for a geometric distribution is $ \frac {1-p}{p^2}$, but I can't seem to relate the formula to the answer above. So I think there maybe some additional formulas or principles related to conditional variance that I miss.

Can someone explain the reasoning for this? Thanks.

$\endgroup$
0
$\begingroup$

The reason: The Geometric Distribution is memoryless. $$\mathsf P(X=x\mid X>4)=\mathsf P(X=x-4)~\mathbf 1_{x>4}$$

Therefore the conditional distribution of $(X-4)$ given $(X>4)$ is the same as the distribution for $X$ ; geometric with variance $6$.

$\endgroup$
  • $\begingroup$ Can you elaborate more about it being memoryless? Perhaps a numeric example or so? Appreciate your help though! $\endgroup$ – speedy_catch Feb 14 at 8:28
  • $\begingroup$ $$\begin{align}\mathsf P(X=x\mid X>4)&=\dfrac{\mathsf P(X=x)~\mathbf 1_{x>4}}{\mathsf P(X>4)}\\&=\dfrac{(1-p)^{x-1}~p~\mathbf 1_{x>4}}{(1-p)^4}\\&=(1-p)^{(x-4)-1}~p~\mathbf 1_{x>4}\\&=\mathsf P(X=x-4)~\mathbf 1_{x>4}\end{align}$$ $\endgroup$ – Graham Kemp Feb 14 at 9:00
  • $\begingroup$ sorry to ask this dumb question, but how do I interpret the $1_x>4$ at the end of the sentences? Thanks so much again! $\endgroup$ – speedy_catch Feb 14 at 9:20
  • 1
    $\begingroup$ @speedy_catch $\mathbf 1_{x>4}$ is an indicator function. $\endgroup$ – callculus Feb 21 at 15:32

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.