Finding an expected value

Question

Below is a problem that I did from Chapter 4 of the book "Intoduction to Probability Theory". The book was written by Hoel, Port and Stone. The answer I got is wrong. I would like to know what I did wrong.
Thanks,
Bob

Problem: Let $X$ be a Poisson with parameter $\lambda$. Compute the mean of $(1+X)^{-1}$.
Answer:
The density function for the Poisson distribution is: $$f(x) = \frac{\lambda^x e ^ {-x}}{x!}$$ Let $u$ be the mean that we seek. \begin{eqnarray*} u &=& \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-x}}{x!((1+x))} = \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-x}}{(x+1)!} \\ u &=& \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-(x-1)}}{(x)!} \\ u &=& \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-x + 1}}{(x)!} \\ u &=& e \sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-x}}{(x)!} \\ \end{eqnarray*} Observe that when $\lambda$ is very large that $u$ is very large. Therefore, I conclude that I am already wrong. The books answer is: $$ \lambda^{-1}(1-e^{-\lambda}) $$

Answer 1

Your $f(x)$ is not correct. The term $e^{-x}$ should be $e^{-\lambda}$.

From your first line to the second, when you offset $x$ by $1$, you should have $\lambda^{x-1}$ in the numerator. That would lead you to pull a factor $\lambda^{-1}$ out of the sum, fixing that part.

Then the final sum would be $1$ if it went from $0$ to $\infty$, so it is $1$ minus the $x=0$ term giving $1-e^{-\lambda}$

Answer 2

\begin{eqnarray*} u &=& \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\color{red}{\lambda}}}{x!((1+x))}\\ &=& \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{(x+1)!} \\ &=& \sum_{x = 1}^{\infty} \frac{\lambda^{x-1} e ^ {-\lambda}}{x!} \\ &=& \frac 1\lambda\sum_{x = 1}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!} \\ &=& \frac 1\lambda\left(\sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!}-e^{-\lambda}\right) \\ &\vdots& \end{eqnarray*}

Answer 3

The Poisson distribution is

$$f(x) = \frac{\lambda^x e ^ {-{\color{red}{\lambda}}}}{x!}, x \in \mathbb{N}. $$

Therefore, the expected value $u$ of $(1+X)^{-1}$ is:

$$u = \sum_{x = 0}^{\infty} \frac{\lambda^x e ^ {-\lambda}}{x!(1+x)} = e ^ {-\lambda}\sum_{x = 0}^{\infty} \frac{\lambda^x}{(1+x)!}.$$

Now, do a substitution $t = x+1$:

$$u = e ^ {-\lambda}\sum_{t = 1}^{\infty} \frac{\lambda^{t-1} }{t!} = \frac{e^{-\lambda}}{\lambda}\sum_{t = 1}^{\infty} \frac{\lambda^{t}}{t!} = \frac{e^{-\lambda}}{\lambda}\left(\sum_{t = 0}^{\infty} \frac{\lambda^{t}}{t!} - 1\right).$$

It is well-known that:

$$\sum_{t = 0}^{\infty} \frac{\lambda^{t} }{t!} = e^{\lambda}.$$

You can check this fact here.

Finally:

$$u = \frac{e^{-\lambda}}{\lambda}(e^{\lambda}-1) = \frac{1}{\lambda}(1-e^{-\lambda}) = \lambda^{-1}(1-e^{-\lambda}).$$