Finding value of $p(x)$ given an MGF

Question

Question:

The MGF of the independent discrete random variable $X$ is given by $$M_X(t) = \left(\frac{1}{2}e^{2t} + \frac{1}{2}e^{4t}\right)^7$$

Find $p_X(15)$

---

I have been staring at this question for a couple of minutes and I don't know how to go about it. I have been trying to find some distribution that works with this MGF, but I can't find one!

I have the solution to the question if anyone is interested, but I don't understand it.

---

Solution:

We have $X = X_1 + . . . + X_7,$ where $P[X_i = 2] = P[X_i = 4] = \frac{1}{2}$.

We have $p_X(15) = 0$, since $X$ can take only even values.

Answer 1

Note that if you have $n$ i.i.d random variables, and $X=X_1+\cdots X_n$, then the M.G.F of $X$ is the product of the M.G.F of the $X_i's$. Since your M.G.F has an exponent of $7$, a good starting guess is that you can "break" your r.v. as a sum of $7$ i.i.d rvs. (That's what your solution has done).

For the solution to be complete, you just need to show that the guess indeed is correct, i.e. is $(\dfrac{1}{2}e^{2t}+\dfrac{1}{2}e^{4t})$ is an M.G.F of some r.v?

The solution again gives you a r.v. for which this is indeed the M.G.F.

The rest is also easy. The r.v. only takes even values, and so the sum of these r.v.s can only take even values. Hence the probability of getting $15$ is $0$

Answer 2

The moment generating function of a sum of independent random variables is the product of the moment generating functions of each variable. When such variables have identical distribution, this means that: $$X=\sum_{i=1}^7 X_i \implies M_X(t) = M_{X_i}(t)^7\\\therefore M_{X_i}(t) = \tfrac 1 2 (e^{2t}+e^{4t}) \qquad \text{for }i\in\{1..7\}$$

That is, the random variable $X$ is the sum of seven i.i.d. random variables with the given moment generating function.

This is the moment generating function of a discrete random variable who takes the values $2$ and $4$ each with a probability of $1/2$. (Can you show?)

Hence: $$\mathsf P(X_i=x) = \begin{cases} 1/2 & :x=2 \\ 1/2 & :x= 4 \\ 0 & :\text{elsewhere}\end{cases}$$

Thus the value of $X$ has a support of $\{14+2n\mid n\in\{0..7\}\}$ or $\{14, 16, ..., 28\}$.

Thus the probability mass at the point $X=15$ is: $p_X(15) = \mathsf P(X=15) = 0$

Answer 3

Another way: the MGF is $ E(e^{Xt})=\sum\limits_x p_X(x) e^{xt}$. By inspection of your MGF, and by the Newton binomial, it is a finite sum of terms of the form $a_k e^{kt}$ - where each $a_k$ would correspond to $p_X(k)$. But the sum has only terms with $k$ even. Hence $a_{15}=p_X(15)=0$