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GATE CS 2017 Set 2 · Engineering Mathematics

Question 48

NATPrinted as CS Q.482 marksNo negative markingEasy (estimate)

Question

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

Official answer and solution
Official answer
54
From
Official GATE 2017 answer key

Solution

Expand the algebraic expression (X+2)2(X + 2)^2 and use the mean and variance properties of the Poisson distribution.

For a Poisson distributed random variable XX with parameter λ=5\lambda = 5:

  • E[X]=λ=5E[X] = \lambda = 5
  • Var(X)=λ=5\text{Var}(X) = \lambda = 5
    Using Var(X)=E[X2]−(E[X])2\text{Var}(X) = E[X^2] - (E[X])^2:
E[X2]=Var(X)+(E[X])2=5+52=5+25=30E[X^2] = \text{Var}(X) + (E[X])^2 = 5 + 5^2 = 5 + 25 = 30

Expand the expectation E[(X+2)2]E[(X + 2)^2] using the linearity of expectation:

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

Substitute E[X2]=30E[X^2] = 30 and E[X]=5E[X] = 5:

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

Thus, the expectation is 54.

Final answer

54

Common trap

Assuming E[(X+2)2]=(E[X]+2)2=(5+2)2=49E[(X+2)^2] = (E[X] + 2)^2 = (5+2)^2 = 49, which ignores the variance of XX.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2017 Computer Science and Information Technology (CS) Session 2
Question
48 of 65, printed as CS Q.48
Section
Computer Science and Information Technology

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