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

76 votes
76 votes
Best answer

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,

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

$\quad = E(X^{2})   +   4 E(X)  +  4$

$\quad =30  +  (4 \times 5)  +  4$

$\quad = 54$

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6 Comments

is above equation is true for normal and uniform distibution?
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how to solve this kind of problems ? can you give any reference please..?
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The pdf of the exponential random variable is given by

$\frac{e^{- \lambda}.\lambda^x}{x!}=f(x)$

The expectation for this is given by $\int_0^{\infty}xf(x)dx$

The expectation of a function of x say $g(x)$ is given by $\int_{0}^{\infty}g(x).f(x)dx$

So,

$E[(x+2)^2]=\int_{0}^{\infty}(x+2)^2f(x)dx=\int_{0}^{\infty}x^2f(x)dx+4\int_{0}^{\infty}xf(x)dx+4\int_{0}^{\infty}f(x)dx$

 

For a continuous random variable defined over interval $[a,b]$, $\int_{a}^{b}f(x)dx=1$

So, $\int_{0}^{\infty}f(x)dx=1$ and here $f(x)$ is our pdf for exponential random variable.

So, $E[(x+2)^2]=E[x^2]+4E[x]+4$
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@

Whatever you just proved is for Continuous Random Variable i,e for exponential distribution

But here we are considering Discrete distribution..i.e poisson distribution.. ??
 

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Do other probabliity distributions also follow 'linearity of expectation'?
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Linearity of expectation is a property of expectation, not specific to any particular probability distribution.
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15 votes
15 votes

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

1 vote
1 vote

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

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–2 votes
–2 votes

a discrete frequency distribution which gives the probability of a number of "independent events" occurring in a fixed time.

E(cX)=cE(x)

E(c)=c  where c is constant

E(XY)=E(X)E(Y)  iff X and Y are independent variable

so here it is given poison distribution and poison distribution is used when event are all distinct.

E(X2+4X+4) = E(X2) + E(4X) + E(4)

=E(X)E(X)+4E(X)+4

=5*5+4*5+4=49 is correct ans

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