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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?

Sri-Amirthan Theivendran
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hedebyhedge
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  • A good start. Then you continue to compute the conditional pmf by definition of conditional probability. – BGM Feb 28 '18 at 07:00

1 Answers1

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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}

Math1000
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  • Nice and clear cut answer. There seems to be a typo for $\mathbb P(X\leqslant Y)$: on the 3rd-to-last line, after the the inner summation, the exoponent for $(1 - p_2)$ should be $k-1$ and not $j-1$. I will remove this comment after the typo is corrected. – Lee David Chung Lin Apr 03 '18 at 05:37