The GATE Grind

GATE 2023 ST – Question 56

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

The joint PMF is e⁻²/[x!(y−x)!] for integers 0≤x≤y,zero otherwise. Which hold?

  1. E(X | Y=4)=2
  2. M_Y(v)=exp[2(e^v−1)] for all real v
  3. E X=2
  4. M_(X,Y)(u,v)=exp[−2+(1+e^u)e^v] for all real u,v

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (A) E(X | Y=4)=2; (B) M_Y(v)=exp[2(e^v−1)] for all real v; (D) M_(X,Y)(u,v)=exp[−2+(1+e^u)e^v] for all real u,v

Explanation

Let Z=Y−X. The joint PMF of X,Z factors as two independent Poisson(1) PMFs. Hence Y=X+Z∼Poisson(2),and X|Y=4∼Binomial(4,1/2),proving A,B. E X=1,so C fails. The joint MGF is E exp[(u+v)X+vZ]=exp(e^(u+v)−1+e^v−1),which is D. **A,B,D**.

More Joint and Conditional Distributions questions

All Joint and Conditional Distributions questions →