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Official PYQ · 2017Set 2 · Q48Probability: Random Variables and DistributionsNAT1 mark

GATE CS 2017 Set 2 Q48 — Probability: Random Variables and Distributions

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If a random variable XX has a Poisson distribution with mean 55, then the expectation E[(x+2)2]E\left [ \left ( x+2 \right )^{2} \right ] equals ___.
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Answer: 54

Given that XX is a Poisson random variable with mean λ=5\lambda = 5.

For a Poisson distribution, both the mean and the variance are equal to λ\lambda:

  • Mean: E[X]=λ=5E[X] = \lambda = 5
  • Variance: Var(X)=λ=5Var(X) = \lambda = 5

Using the relation between variance and expectation:

Var(X)=E[X2]−(E[X])2Var(X) = E[X^2] - (E[X])^25=E[X2]−525 = E[X^2] - 5^2E[X2]=5+25=30E[X^2] = 5 + 25 = 30

We need to find the expectation E[(X+2)2]E[(X+2)^2]. Expanding the expression inside the expectation:

E[(X+2)2]=E[X2+4X+4]E[(X+2)^2] = E[X^2 + 4X + 4]

By the linearity of expectation:

E[(X+2)2]=E[X2]+4E[X]+4E[(X+2)^2] = E[X^2] + 4E[X] + 4

Substituting the values of E[X2]E[X^2] and E[X]E[X]:

E[(X+2)2]=30+4(5)+4=30+20+4=54E[(X+2)^2] = 30 + 4(5) + 4 = 30 + 20 + 4 = 54

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