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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 ___.

edited
I think 54 should be answer.
$x^{2} + 4x + 4$

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

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:-

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

$E(constant) = Constant.$

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

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

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

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,

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

$\quad = E(X^{2}) + 4 E(X) + 4$

$\quad =30 + (4 \times 5) + 4$

$\quad = 54$

is above equation is true for normal and uniform distibution?
how to solve this kind of problems ? can you give any reference please..?
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, $E[(x+2)^2]=E[x^2]+4E[x]+4$

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.. ??

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

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

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

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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