I have proved that $$E[X^n]=\lambda E[(X+1)^{n-1}]$$ for a Poisson random variable with parameter $\lambda$ where $n$ is a positive integer.
Now $E[X+2]=E[X]+2=\lambda+2$
But using the formula that I proved,
$E[X+2]=\lambda E[(X+3)^0]=\lambda E[1]=\lambda$
Which of these is correct and where have I gone wrong?
Proof of $E[X^n]=\lambda E[(X+1)^{n-1}]$:
\begin{align} E[X^n]&=\sum_{i=0}^\infty i^np(i) \\ &=\sum_{i=0}^\infty i^ne^{-\lambda}\frac{\lambda^i}{i!} \\ &=\lambda\sum_{i=0}^\infty i^ne^{-\lambda}\frac{\lambda^{i-1}}{i!} \\ &=\lambda\sum_{i=1}^\infty i^ne^{-\lambda}\frac{\lambda^{i-1}}{i!} \\ &=\lambda\sum_{i=1}^\infty i^{n-1}e^{-\lambda}\frac{\lambda^{i-1}}{(i-1)!} \\ &=\lambda\sum_{i=0}^\infty(i+1)^{n-1}e^{-\lambda}\frac{\lambda^i}{i!} \\ &=\lambda\sum_{i=0}^\infty(i+1)^{n-1}p(i) \\ &=\lambda E[(X+1)^{n-1}] \end{align}
$\endgroup$ 23 Answers
$\begingroup$$X \sim \text{Poisson}(\lambda)$ and $X+2$ have different CDFs (see here), so as Seyhmus said, $X+2$ is not Poisson.
$\endgroup$ $\begingroup$But using the formula that I proved,
No. The formula applies to a Poisson random variable. Here: $X$ is Poisson, but $(X+2)$ is not. You can use the formula for $\mathsf E(X^n)$ but it is inapplicable to $\mathsf E((X+2)^n)$.
Properly applying your fomula, and the linearity of expectation, gives:
$\mathsf E(X+2) ~{= \mathsf E(X^1)+2 \\= \lambda\, \mathsf E((X+1)^0)+2 \\= \lambda +2}$
As it should.
Edit: The proof, that which you just posted, checks out okay.
$\endgroup$ $\begingroup$It looks as if you're trying to apply a simple rule of logic that says if something is true of all values of $X$, then it is true of $X+2$.
But the statement that $\operatorname{E}(X^n) = \lambda \operatorname{E}((X+1)^{n-1})$ does not say something is true of all values of $X$. Rather, it is true of random variables with a particular distribution. It can be applied to random variables with that distribution, but not to other things. The "particular distribution" in question assigns a non-zero probability to each of $0,1,2,3,\ldots\ .$ The distribution of the random variable $X+2$ does not do that; it assigns nonzero probability to each of $2,3,4,\ldots$ and zero probability to $0$ and to $1.$ It does not have the distribution to which the proposition you proved is applicable.
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