# Geometric PMF and the variance

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.

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 $$6$$.
• \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} – Graham Kemp Feb 14 at 9:00
• 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! – speedy_catch Feb 14 at 9:20
• @speedy_catch $\mathbf 1_{x>4}$ is an indicator function. – callculus Feb 21 at 15:32