Variance of Poisson distribution multiplied by a constant

Question

Assume I have a Poisson distribution $x=Poisson(X)$. It is clear that the mean and the variance will be equal to X.

In the case we have of $B *Poisson(X)$ being B a constant, I would like to see a small proof showing why the variance is $B^2*X$ and the mean is $B*X$

Many thanks!! :)

Answer 1

I'm a bit confused by your notation but there's nothing unique about the Poisson in this regard.

Say you have a random variable X, with mean E(X) and variance V(X)

Then let $Y=cX$, for some constant c.

$\displaystyle{E(Y)=\sum_{y} y P(Y=y)=\sum_{x} cx P(cX=cx)=\sum_{x} cx P(X=x)= c\sum_{x} P(X=x)=cE(X)}$

Then $V(Y) = E(Y^2) - (E(Y))^2 = E(c^2X^2) - c^2(E(X))^2 = c^2(E(X^2)-(E(X))^2) = c^2V(X)$

Answer 2

Hint: Makes use of the rules $\operatorname{E}[aX]=a\operatorname{E}[X]$ and $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. Where $X$ is a random variable and $a \in \mathbb{R}$.

Answer 3

@Maru_kai: If you want to prove the rules described in the previous answer you can easily do it by yourself

$$\mathbb{E}[aX]=\sum_{i}a x_i p(x_i)=a\sum_{i} x_i p(x_i)=a\mathbb{E}[X]$$

try by yourself to prove that $\mathbb{V}[aX]=a^2\mathbb{V}[X]$ by using the definition

$\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}^2[X]$