Krdytkyu

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:
$$ lambda^{-1}(1-e^{-lambda}) $$

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 $t = x+1$:

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

It is well-known that:

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

You can check this fact here.

Finally:

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

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 $1-e^{-lambda}$

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 1lambdasum_{x = 1}^{infty} frac{lambda^x e ^ {-lambda}}{x!}
\ &=& frac 1lambdaleft(sum_{x = 0}^{infty} frac{lambda^x e ^ {-lambda}}{x!}-e^{-lambda}right) \ &vdots&
end{eqnarray*}