GATE CSE 2017 | Set 2 | Question: 48

Question

If a random variable $X$ has a Poisson distribution with mean $5$, then the expectation $E\left [ \left ( x+2 \right )^{2} \right ]$ equals ___.

Comments

I think 54 should be answer.

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$x^{2} + 4x + 4$

for poisson, mean and variance are same.

E($x^{2}$) – 25 = 5

E($x^{2} + 4x + 4$) = E($x^{2}$) + E($4x$) + E($4$)

you will get 54

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Important Theorem in Poisson Distribution:-

$Mean = Variance = E(X) =$ $\lambda$ = $np$ where, n is very large and p is very small so that $np$ is finite.

General Theorems:-

$\text{Variance}  = E\left(X^{2}\right) -  [ E(X) ]^{2}$   

$E(constant) = Constant.$

$E(aX+b) = aE(X) + E(b) \implies aE(X) + b$

$Variance(constant) = \sigma^2(X) =  0$

$Variance(aX+b) = a^2 Var(X)$

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Ans

Answer 1

In Poisson distribution :

Mean  =  Variance  as n is large and p is small

And we know:

$\text{Variance}  =  E\left(X^{2}\right)  -   [ E(X) ]^{2}$  

$\Rightarrow E(X^{2})     =  [ E(X) ]^{2}  + \text{Variance}$

$\Rightarrow E(X^{2})     =   5^{2}   + 5$

$\Rightarrow E(X^{2})     =   30$

So, by linearity of expectation,

$E[(X + 2)^{2} ]   =   E[ X^{2} + 4X + 4 ]$

$\quad = E(X^{2})   +   4 E(X)  +  4$

$\quad =30  +  (4 \times 5)  +  4$

$\quad = 54$

Comments

Linearity of expectation is a property of expectation, not specific to any particular probability distribution.

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for binomial distribution, MEAN  = np , variance = npq , where n is the no of trials .

for poisson distribution MEAN = VARIANCE = lamda

 

 

var = e(x^2) - [ e(x) ] ^2

as var = mean in case of poisson dis,

5 = e(x^2) - 25

thus, e(x^2) = 30

 Variance of a constant is 0 while expectation of constant is constant as expectation is avg whereas variance is the change.

E[(x+2)^2]  = e(x^2) + 4e(x) + e(4)  = 30 + 4 * 5 + 4 = 54

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thank you

Answer 2

E(X²+4X+4) = E(X²) + E(4X) + E(4)

we know that V(x) = E(x²) - (E(x))²   =    5 = E(x²) - 25        =>  E(x²) = 30

So      30 + 4*5 + 4  =  54

Answer 3

In Poisson distribution of a random variable X, $E[X] = Variance[X] = \lambda$

$E[X]=5\; (Given)$

$E[X+2]=E[X]+E[2]=5+2=7$

$Variance[X+2]=Variance[X]+Variance[2]=5+0=5$

$\Rightarrow Variance[X+2] = E[{(X+2)}^{2}] - {(E[X+2])}^{2}$

$\Rightarrow 5 = E[{(X+2)}^{2}] - {7}^{2}$

$\Rightarrow E[{(X+2)}^{2}] = 54$

Answer 4

Ans

Answer 5

As it is given in the question that mean = 5 

so,we have E[x] = Y = 5 and VAR[x] = Y = 5

Now, E [ (x+2)^2 ] = E [ x^2 + 4x + 4 ]

                            = E [x^2] + 4E [ x ] + 4

                             =E [x^2] + 20 + 4

we can calculate E [x^2] using VAR [x] 

VAR [ x ] = E [x^2] - (E [x])^2

so,E [x^2] =  VAR [ x ] + (E [x])^2

                 = 5 + 25

                 = 30

Finally ,

E [ (x+2)^2 ] = E [x^2] + 20 + 4

                     = 30 + 20 + 4

                     =54

E [ (x+2)^2 ] = 54

Answer 6

Answer is 54

Expectation of constant remain constant.

E(X)=V(X)=5 (given)

V(X)=E(X^2)-(E(X)^2)

5=E(X^2)-25

E(X^2)=30

 

Expectationn-and-variance-both-equal-to-lambda-for-a-poisson-distribution

https://gateoverflow.in/?qa=blob&qa_blobid=610850049674317470