Let $X \sim \text{Geo}(p_1)$ and $Y \sim\text{Geo}(p_2)$. What is $E(X\mid X \leq Y)$?

Question

I'm not sure how to solve this question, but my intuition would be that we would sum up

$\sum_{x=1}^{\infty} (x * P(X=x\mid X \leq Y))$, but I would still be stick with a conditional probability.

Is there a general approach to solving conditional expectations between two different distributions?

Answer 1

Using the definition of conditional probability, \begin{align} \mathbb E[X\mid X\leqslant Y] &= \sum_{k=1}^\infty k\mathbb P(X=k\mid X\leqslant Y)\\ &= \sum_{k=1}^\infty k\frac{\mathbb P(\{X=k\}\cap \{X\leqslant Y\})}{\mathbb P(X\leqslant Y)}. \end{align} Assuming $X$ and $Y$ are independent, we compute \begin{align} \mathbb P(X\leqslant Y)&= \sum_{k=1}^\infty \mathbb P(\{ X\leqslant Y\}\cap\{X=k\})\\ &= \sum_{k=1}^\infty \mathbb P(X=k) \sum_{j=k}^\infty \mathbb P(Y=j)\\ &= \sum_{k=1}^\infty (1-p_1)^{k-1}p_1 \sum_{j=k}^\infty (1-p_2)^{j-1}p_2\\ &= \sum_{k=1}^\infty (1-p_1)^{k-1}p_1 (1-p_2)^{j-1}\\ &= p_1\sum_{k=1}^\infty ((1-p_1)(1-p_2))^{k-1}\\ &= \frac{p_1}{1-(1-p_1)(1-p_2)}. \end{align} It follows that \begin{align} \mathbb E[X\mid X\leqslant Y] &= \frac{1-(1-p_1)(1-p_2)}{p_1}\sum_{k=1}^\infty k (1-p_1)^{k-1}p_1\sum_{j=k}^\infty (1-p_2)^{j-1}p_2\\ &=(1-(1-p_1)(1-p_2))\sum_{k=1}^\infty k (1-p_1)^{k-1}(1-p_2)^{k-1}\\ &=(1-(1-p_1)(1-p_2)) \sum_{k=0}^\infty (k+1) ((1-p_1)(1-p_2))^k\\ &= \frac1{1-(1-p_1)(1-p_2)}. \end{align}