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I am struggling with the following conditional probability problem.

Let $G \sim \mathrm{Geometric}(p)$, $U \sim \mathrm{Unif} \{1, \ldots ,n\}$, and $G$ & $U$ be independent. Determine the conditional probability $\mathrm{Pr}(G=U \mid G \le n)$.

Using the definition of conditional probability we obtain

$$\mathrm{Pr}(G=U \mid G \le n) = \frac{\mathrm{Pr}(G=U, G \le n)}{Pr(G \le n)}$$

The denominator is easy enough to compute,

$$\mathrm{Pr}(G \le n) = \sum_{i=1}^{n}p(1-p)^{i-1}$$

As for the numerator, I believe we have

$$\mathrm{Pr}(G=U, G \le n) = \sum_{i=1}^{n}\left(p(1-p)^{i-1} \cdot \frac{1}{n}\right) = \frac{1}{n} \sum_{i=1}^{n}p(1-p)^{i-1},$$

so that the final answer is $\frac{1}{n}$. Is this correct?

rubik
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wacko_wanderer
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    Your result is correct, but whenever you write "I believe", you are obviosly missing some steps to be convinced yourself. So you should be more detailed to show that $$\mathrm{Pr}(G=U, G \le n) = \sum_{i=1}^{n}\left(p(1-p)^{i-1} \cdot \frac{1}{n}\right)$$ really holds. Hint: ${G \le n} = \bigcup_{k=1}^n {G=k}$ – Gono Dec 15 '22 at 09:12

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