Gate2017 ss Q4

0. If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X + 2)2] equals _____

  • Option : A
  • Explanation :
    Using Linearity of Expectation, we can write,
    E[(X+2)2] = E[X2] + E[4X] + E[4]
    The Poisson distribution, mean and variance are same. Here Mean is given as 5. So variance should also be 5.
    Also,
    Variance = E[X2] – (E[X])2
    5 = E[X2] – 25.
    E[X2] = 30
    Thus E[(X+2)2] = 30 + 4*5 + 4 = 54.
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