Given a random variable X, prove f(x) is a probability function.

Question

Let X be a random variable with $f_X(x) = P(X = x) = \frac{e^{-\mu}\mu^{x}}{x!}$ for $x \geq 0$

Prove that $f_X(x)$ is a probability mass function. Prove that $P(X$ is even $) = \frac{1}{2}(1 + e^{-2\mu})$

Work done so far:

First of all I assume there was a typo in the question since the definition of a probability function includes the following condition: $P(X = x) = f_X(x) = f_X$

I believe I proved that by trivially using the definition of $X$ that was given.

However, it must also fulfill the condition that: $\forall x \in R, f_X(x) \geq 0$ and that $f_X$ sums to 1.

I don't know how to prove that first of these part, since I don't know what $\mu$ is equal to. I read that $\mu$ is the expected value of $X$, but I also don't know how to calculate that without using the function that $\mu$ is part of. Could anyone give me any pointers?

Answer 1

The distribution you are citing is called Poisson distribution with parameter $\mu$ and write $X\sim$ Poiss($\mu$). Now we want to show, that $f_X(x) = \mathbf{P}(X = x) = \frac{e^{-\mu}\mu^{x}}{x!}$ is really a pmf with mass on the non-negative integers. We don't need to know what $\mu$ actually describes - as long as $\mu>0$ holds as pointed out in the comment - we have \begin{align} \sum_{x=0}^{\infty}\mathbf{P}(X = x)=\sum_{x=0}^{\infty}\frac{e^{-\mu}\mu^{x}}{x!}=e^{-\mu}\sum_{x=0}^{\infty}\frac{\mu^{x}}{x!}=e^{-\mu}e^{\mu}\equiv 1 \end{align} holds for all real $\mu$ but we have $\mathbf{P}(X = x)\geq 0$ only for $\mu>0$. As a matter of fact, $\mu$ inded describes the expectation of a Poisson distributed random variable.

Now let's try to figure out, what the probability of a Poisson distributed random variable $X$ is only taking even realizations. So we want to know \begin{align} \mathbf{P}(X \text{ "is even"})=\sum_{x=0}^{\infty}\mathbf{P}(X = 2x)=\sum_{x=0}^{\infty}\frac{e^{-\mu}\mu^{2x}}{(2x)!} \tag 1 \end{align} Now there several possibilities to determine the value of $(1)$. For example we know that \begin{align} \mathbf{P}(X \text{ "is even"})+\mathbf{P}(X \text{ "is odd"})=1 \\ \iff \\ \sum_{x=0}^{\infty}\frac{e^{-\mu}\mu^{2x}}{(2x)!}+\sum_{x=0}^{\infty}\frac{e^{-\mu}\mu^{2x+1}}{(2x+1)!}=1 \end{align} and further we can see that (needs to be justified) $\frac{d}{d\mu}\sum_{x=0}^{\infty}\frac{\mu^{2x+1}}{(2x+1)!}=\sum_{x=0}^{\infty}\frac{\mu^{2x}}{(2x)!}$ which eventually leads to a linear ODE which could we now solve.

However, the easiest way is to check some table and then we can see that indeed we have \begin{align} \sum_{x=0}^{\infty}\frac{\mu^{2x}}{(2x)!}=\cosh(\mu)=\frac{e^{\mu}+e^{-\mu}}{2} \tag 2 \end{align} and therefore by using $(2)$ \begin{align} e^{-\mu}\sum_{x=0}^{\infty}\frac{\mu^{2x}}{(2x)!}=e^{-\mu}\cosh(\mu)=e^{-\mu}\frac{e^{\mu}+e^{-\mu}}{2}=\frac12(1+e^{-2\mu}) \tag 3 \end{align} Altogether we have because of $(3)$ \begin{align} \mathbf{P}(X \text{ "is even"})=\sum_{x=0}^{\infty}\mathbf{P}(X = 2x)=\sum_{x=0}^{\infty}\frac{e^{-\mu}\mu^{2x}}{(2x)!}=\frac12(1+e^{-2\mu}) \end{align} which is exactly what we wanted to show.

Answer 2

This is not an answer but it got too long to be a comment.

$\mu$ is a positive number. Also, I don't know what your book means by a "probability function" because that phrase is not used in English. Are you translating from another language? In English, the probability mass function of a discrete random variable is denoted by $p_X(\cdot)$ (sometimes as $f_X(\cdot))$ and $p_X(a)$ is the probability that $X = a$. $p_X(a)$ has positive value only for some values of $a$, not the $x \geq 0$ meaning all nonnegative real numbers that you seemingly assert in your first sentence. Also, your definition of probability function should include a statement such as $\sum_a p_X(a) = 1$ (where the sum is restricted to those $a$ for which $p_X(a) > 0$) in addition to $p_X(a) \geq 0$. If you did not check the former condition, how can you assert that what you have is a "probability function"?