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

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2 votes

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.