Question

Let \(X\) and \(Y\) be independent and identically distributed geometric random variables having the following probability mass function
\[ P(X=x)=p(1-p)^x, \quad x=0,1,2,\ldots, \]
where \(p\in(0,1)\). Then which of the following statements is correct?

• A. \(E(X|X+Y)=\dfrac{X+Y}{2}\)
• B. \(P(X=Y)=1\)
• C. \(P(X=Y)=\dfrac{1-p}{1+p}\)
• D. \(E(X|X+Y)=\dfrac{X+Y+2}{2}\)

Hint:

Since \(X\) and \(Y\) are i.i.d., swapping their labels does not change the joint distribution, so \(E(X|X+Y)\) and \(E(Y|X+Y)\) must be equal and together add up to \(X+Y\).

Answer 1

The Correct Option is A

Step 1: Set up the conditional distribution.
$X$ and $Y$ are iid with $P(X=x)=p(1-p)^x$. We want $E(X \mid X+Y=n)$ for a fixed total $n$.

Step 2: Find the conditional pmf directly.
For $0 \le k \le n$,
\[ P(X=k, Y=n-k) = p(1-p)^k \cdot p(1-p)^{n-k} = p^2(1-p)^n. \]
This does not depend on $k$: every split of the total $n$ has the same joint probability.

Step 3: Normalize to get the conditional pmf.
There are $n+1$ possible values of $k$, so
\[ P(X=k \mid X+Y=n) = \frac{1}{n+1}, \quad k=0,1,\ldots,n. \]
So given the total is $n$, $X$ is exactly uniform over $\{0,1,\ldots,n\}$.

Step 4: Compute the conditional mean.
\[ E(X\mid X+Y=n) = \frac{0+1+\cdots+n}{n+1} = \frac{n}{2}. \]

Step 5: Write this back in terms of $X+Y$.
\[ E(X\mid X+Y) = \frac{X+Y}{2}. \]
Matches option (A), rules out (D).

Step 6: Rule out (B) and (C).
$X$ is uniform on $\{0,\ldots,n\}$ given the sum, not always equal to $Y$, so (B) fails. Direct summation gives $P(X=Y)=p/(2-p)$, not matching (C).

Final Answer:
\[ \boxed{E(X\mid X+Y)=\dfrac{X+Y}{2}} \]