Let $X$ be Poisson random variable with mean $1$. Find $E(|X-1|)$

Question

Let $X$ be Poisson random variable with mean $1$. Find $E(|X-1|)$

Frankly I have no idea how to attempt this one. Help will be much appreciated

Answer 1

$E|X-1|=P(X=0)+\sum_{n=1}^{\infty}(n-1)P({\{X=n\})}$. Can you do the computation now?

Answer 2

The mean absolute deviation about the mean of Poisson distribution is:

$$\mathbb{E}[|X-\lambda|]=2e^{-\lambda}\dfrac{\lambda^{\lfloor\lambda\rfloor+1}}{\lfloor\lambda\rfloor!}$$

where $\lfloor\lambda\rfloor$ is the floor function.

Therefore,

$$\mathbb{E}[|X-1|]=2e^{-1}$$

Answer 3

The answers by Kavi Rama Murthy and QFi already give you the answer, I'll just present another approach, just for fun.

Since $$ |X-1| = \left\{ \begin{array}{cc} 2 + X - 1 & \text{if} \ X = 0 \\ X - 1 & \text{if} \ X \geq 1 \end{array} \right. $$

We have that $$ E(| X - 1 |) = 2 P(X=0) + E(X - 1) $$

And so $$ E(| X - 1 |) = 2 P(X=0) + E(X) - E(1) $$ Since $E(X) = \lambda = 1$ and $E(k) = k$, for any constant $k$, $$ E(| X - 1 |) = 2 e^{-1}\frac{1^0}{0!} + 1 - 1 $$ Simplifying gives $$ E(| X - 1 |) = 2 e^{-1} $$

Answer 4

Note that $$ |k-1| = 1, \text{ for } k=0 \text{ w.p. } P(X=0)\\ |k-1| = 0, \text{ for } k=1 \text{ w.p. } P(X=1)\\ |k-1| = k-1, \text{ for } k \ge 1 \text{ w.p. } P(X=k) $$

Then the expectation $E|X-1|$ becomes: $1\times e^{-\lambda}+0\times \lambda e^{-\lambda}+ \sum_{k=1}^\infty (k-1)e^{-\lambda} \lambda^ k /k!$.

Splitting the sum in two parts we get $E|X-1| = \lambda -1 + 2e^{-\lambda}.$

Answer 5

Hint: $\lower{2ex}{\begin{align}\mathsf E(\lvert X-1\rvert)&=\mathsf E(\mathbf 1_{X=0})+\mathsf E((X-1)\mathbf 1_{X\geq 1})\\&=\mathsf P(X{=}0)+\mathsf E(X\mathbf 1_{X\geq 1})-\mathsf P(X{\ge}1)\end{align}}$