# How to find probability distribution function given the Moment Generating Function

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

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.

• Usually you use the fact that the MFG uniquely characterizes the distribution of $X$. For example, $M_{X}(t)$ is the MGF of the Poisson distribution for $\lambda = 2$, hence the probably mass function must be $2^{k}e^{-2}/k!$, $k\in \{0,1,2\dots\}$. Dec 6, 2013 at 10:58
• @JessicaK: Still, is it too lengthy a calculation to find $f_x(x)$ with the brute-force approach I proposed in the question? I'd like to find a method to derive your quick observation that $M_x(t)$ is the MGF of a Poission distribution for $\lambda = 2$.
– NaN
Dec 6, 2013 at 11:01
• Well, you are not working on a continuous distribution for one, this is an infinite sum (also the X in your integrand should also be lower case). I don't believe it is possible to recover $f_{X}(x)$ without prior knowledge of other MGFs with just the information you have available. Dec 6, 2013 at 11:09
• @JessicaK, what additional information provided would allow you to derive $M_X(t)$ from the equation tagged (1)? Dec 6, 2013 at 11:20
• There seems to be a typo in your definition of $M_Y(t)$ - there's no occurence of $t$ on the right-hand side there...
– fgp
Dec 6, 2013 at 11:49

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

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

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