If $X$ be a Poisson distributed random variable with the parameter $\lambda$, then prove that $E(|X - 1|) = \dfrac{2\sigma} {e}$
$\sigma$ here means the square root of varians (recall that $\sigma^2$ is varians symbol). Hope this provide true information, because the problem doesn't really explicitly said what is sigma here means.
Please help me this one. I am in trouble with the absolute value things. Someone gave me this clue, but it took me to the end of the horizon
It's not complicated if you realize this: $|X-1|=X-1$ if $x \geq 1$ and $1$ if $x = 0$.
Yeah, we should define the absolute value function like that, now I think that it doesn't lead me to prove what should I prove later. And how?
If I take $\lambda=1$,
\begin{align}E(|X-1|)&=\frac{1}{e}\sum_{j=0}^\infty|j-1|\frac{1}{j!}\\&=\frac{1}{e}\left[1+\frac{1}{2!}+2\cdot\frac{1}{3!}+3\cdot\frac{1}{4!}+4\cdot\frac{1}{5!}+\cdots\right]\\&=\frac{1}{e}\left[1+\frac{1}{2!}+(3-1)\frac{1}{3!}+(4-1)\frac{1}{4!}+(5-1)\frac{1}{5!}+\cdots\right]\\&=\frac{1}{e}\left[\left(1+\frac{1}{2!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)-\left(\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots\right)\right]\\&=\frac{1}{e}(1+1)=\frac{2}{e}\times\text{sd}\end{align}
Now you can try to do this for any $\lambda$. For $\lambda=1$, the result is not surprising since the mean absolute deviation about mean for a $\text{P}(\lambda)$ distribution is $\displaystyle E(|X-\lambda|)=2e^{-\lambda}\frac{\lambda^{\lfloor\lambda\rfloor+1}}{(\lfloor\lambda\rfloor)!}$.