610 views

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

1. $\lambda$
2. $2 \lambda$
3. $\lambda^2$
4. $4 \lambda^2$

### 1 comment

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

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.

by
Ans is option (B).