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GATE 2026 ST – Question 41

Joint Distributions · Joint and Conditional Distributions · 2 marks · Multiple choice

Independent iid geometric X,Y have P(X=x)=p(1−p)^x,x=0,1,…,0<p<1. Which holds?

  1. E(X|X+Y)=(X+Y)/2
  2. P(X=Y)=1
  3. P(X=Y)=(1−p)/(1+p)
  4. E(X|X+Y)=(X+Y+2)/2

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Show answer and explanation

Correct answer: (A) E(X|X+Y)=(X+Y)/2

Explanation

Given X+Y=s, each x=0,…,s has the same probability proportional to p²(1−p)^s, so X is conditionally uniform and has mean s/2. Thus **A**. Also P(X=Y)=p²Σ(1−p)^(2k)=p/(2−p), refuting B,C.

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