Let X be a Poisson random variable, show that $E[X^n] = \lambda E[(X+1)^{n-1}]$

Question

I am struggling to understand some of the steps of the proof of this problem:

Let X be a Poisson random variable, show that $E[X^n] = \lambda E[(X+1)^{n-1}]$.

The proof: $$ \begin{align*} E(X^n) &\stackrel{(1)}= e^{-\lambda} \sum_{j=0}^\infty j^n \frac{\lambda^j}{j!} \\ &\stackrel{(2)}= e^{-\lambda} \sum_{j=0}^\infty j^n \frac{\lambda^j}{j!} \\ &\stackrel{(3)}= e^{-\lambda} \sum_{j=1}^\infty j^{n-1} \frac{\lambda^j}{(j-1)!} \\ &\stackrel{(4)}= \lambda e^{-\lambda} \sum_{j=1}^\infty j^{n-1} \frac{\lambda^{j-1}}{(j-1)!} \\ &\stackrel{(5)}= \lambda e^{-\lambda} \left( \sum_{i=0}^\infty (i+1)^{n-1} \frac{\lambda^i}{i!} \right) \\ &\stackrel{(6)}= \lambda E((X+1)^{n-1}). \end{align*} $$

Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):

1. This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0\leq i\leq \infty $ (hope my questions was clear)

2. Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1

3. This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me

4. Nothing special here

5. We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?

6. Why is this conclusion correct?

Thanks in advance!

Answer 1

$1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $\sum\limits_x f(x)\, \mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $\mathbb P(X=x) = e^{-\lambda}\dfrac{\lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers

$3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $\dfrac{j^n}{j!} = \dfrac{j^{n-1}}{(j-1)!}$

$5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $\lambda^{j-1}=\lambda^i$ and $(j-1)!=i!$ and $j=1 \implies i=0$

$6$ - this is the reverse of the first point: you have inside the brackets $\sum\limits_x g(x)\, \mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$

Answer 2

1. This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.
2. Sure.
3. You're simply cancelling the $j$ from $j^n$ with the $j$ from $\frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.
4. Okay.
5. This is a basic change of index. The basic form for this is $\sum_{k=2}^{\infty}k = \sum_{j=1}^{\infty}(j+1)$.
6. We substitute line 5 into the original form to get this result.