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: Understand what is being asked.
\(X\) and \(Y\) are independent and identically distributed (i.i.d.) geometric random variables with pmf \(P(X=x)=p(1-p)^x\) for \(x=0,1,2,\ldots\). We need to find \(E(X|X+Y)\), the conditional expectation of \(X\) given the sum \(X+Y\).

Step 2: Use the symmetry argument for i.i.d. random variables.
Since \(X\) and \(Y\) are i.i.d., the pair \((X,Y)\) and the pair \((Y,X)\) have exactly the same joint distribution. This means that for any fixed value of the sum \(X+Y=n\), \(X\) and \(Y\) play completely symmetric roles.

Step 3: Write the conditional expectation identity.
By linearity of conditional expectation,
\[ E(X|X+Y) + E(Y|X+Y) = E(X+Y|X+Y) = X+Y. \]
By the symmetry noted in Step 2, \(E(X|X+Y)\) and \(E(Y|X+Y)\) must be equal, since swapping the labels \(X\) and \(Y\) does not change the joint distribution or the value of the sum. So
\[ E(X|X+Y)=E(Y|X+Y). \]

Step 4: Solve for the conditional expectation.
Substituting \(E(Y|X+Y)=E(X|X+Y)\) into the sum identity gives
\[ 2E(X|X+Y)=X+Y, \]
so
\[ E(X|X+Y)=\frac{X+Y}{2}. \]

Step 5: Check option (B).
\(P(X=Y)=1\) would mean \(X\) and \(Y\) are equal with certainty, but they are two separate random draws that can differ, so this is false.

Step 6: Check option (C).
Directly, \[ P(X=Y)=\sum_{x=0}^{\infty}[p(1-p)^x]^2 = \frac{p^2}{1-(1-p)^2} = \frac{p}{2-p}. \] This is not the same as \(\dfrac{1-p}{1+p}\) in general, so (C) is false.

Step 7: Check option (D).
There is no reason for an extra constant \(2\) to appear in the numerator; the exact split is \(\dfrac{X+Y}{2}\) with no added term, so (D) is false.

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