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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 ___.

0

**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:-**

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

$E(constant) = Constant.$

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

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

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

2

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,

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

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

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

$\quad = 54$

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

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

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

9

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

–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(X^{2}+4X+4) = E(X^{2}) + E(4X) + E(4)

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

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