ISI2016-MMA-9

Question

Suppose $X$ and $Y$ are two independent random variables both following Poisson distribution with parameter $\lambda$. What is the value of $E(X-Y)^2$ ?

• A. $\lambda$
• B. $2 \lambda$
• C. $\lambda^2$
• D. $4 \lambda^2$

Comments

Its solution is exactly same as https://gateoverflow.in/118513/gate2017-2-48

Answer 1

In Poisson distribution :

Mean  =  Variance  as n is large and p is small

And we know:

   $Variance=E(X^2)−[E(X)]^2 $

  $⇒E(X^2)=[E(X)]^2+Variance = λ^2 + λ$

  Similarly, $ E(Y^2)=[E(Y)]^2+Variance = λ^2 + λ$

$ E[2XY] = 2 * E[X] * E[Y] $

 $ E(X−Y)^2 = E[X^2+2XY+Y^2] = E[X^2] + E[2XY]  + E[Y^2]$ 

 $= λ^2 + λ - 2.λ.λ+ λ^2 + λ) $

$ = 2λ$

So B is correct.

Ref: https://gateoverflow.in/118513/gate2017-2-48

Answer 2

Ans is option (B).