How to find probability distribution function given the Moment Generating Function

Question

After searching, I found two questions like mine, but didn't see my answer to my question.

1. Finding a probability distribution given the moment generating function
2. Finding probability using moment-generating functions

My question is how to find any probability distribution function, given its moment generating function. In particular, how to find this from First Principles (and not memorizing a table).

Let's try an example:

Let $ X \perp Y$. Define the moment generating functions for $X, Y$ respectively as $$M_X(t)=\exp(2e^t-2), M_Y(t)=\left(\frac{3}{4}e^t+ \frac{1}{4}\right)^{10}$$ Find $P(X+Y = 2)$.

First, the problem doesn't tell us whether the distributions are continuous or discrete, so I assume continuous. Now, how do we solve the following for $f_X(x)$?

$$M_X(t)= \int_{-\infty}^{\infty}e^{xt} f_X(x) \ dx = \exp ( 2\ e^t - 2)\tag{1}$$

Next, can we take the derivative with respect to $x$ to both sides, to bring us closer to the solution $f_X(x)$?

I read that a m.g.f. $m_X(t)$ is characteristic to and unique to the distribution of $X$. I saw something about Laplace Transforms in another question, but we have learned nothing of that sort in this course.

Answer 1

There is not any short method to find out probability function but you can analyse the MGF given to you to the MGF known to you of some known probability distribution

so now lets analyse it

$$MGF \quad of \quad Poisson(\lambda)\quad is \quad exp(\lambda(e^t-1))$$ so by matching it to the MGF of X we will understand that

$$X \sim Poi(2)$$

similarly $$MGF \quad of \quad Binomial(n,p)\quad is \quad (q+pe^t)^n$$

so by matching it to the MGF of Y we will understand that $$Y \sim Bin(10, \frac {3} {4})$$

so we want $P(X+Y=2)$

So $$P(Z=2)=P(X=0,Y=2)+P(X=1,Y=1)+P(X=2,Y=0)$$ where $Z=X+Y$

now we know X and Y are independent

$$P(Z=2)=P(X=0)P(Y=2)+P(X=1)P(Y=1)+P(X=2)P(Y=0)$$ $$P(Z=2)=\bigg[ \frac{2^0e^{-2}}{0!}\times \binom{2}{2} {\bigg(\frac{3}{4}\bigg)}^2 {\bigg(\frac{1}{4}\bigg)}^0 \bigg] + \bigg[ \frac{2^1e^{-2}}{1!}\times \binom{2}{1} {\bigg(\frac{3}{4}\bigg)}^1 {\bigg(\frac{1}{4}\bigg)}^1\bigg] + \bigg[ \frac{2^2e^{-2}}{2!}\times \binom{2}{0} {\bigg(\frac{3}{4}\bigg)}^0 {\bigg(\frac{1}{4}\bigg)}^2\bigg] $$

$$P(X+Y=2)= \frac{23e^{-2}}{16} \approx 0.1945$$