It is known that moment generating function generates moments, but does it hold information about the probability of the random variable being realised at a particular value?


Focusing on only discrete random variable,

we have $$M_Y(t) = \frac{2^t+e^{3t}+4k}{3}$$ for $t \in \mathbb{R}$ and $k$ is a constant.

Question For the moment generating function in the example, how is it possible we know $\text{Pr}(Y=1)$ without comparing to the known form of some famous probability mass function?


4 Answers 4


Note that with a discrete distribution the contribution of an atom of probability at some particular $x$-value $x_0$ has a characteristic effect on the mgf.

For example, the probability $p_0$ at $X=0$ will have a term in the mgf of $p_0$.

A probability of $p_1$ at $X=1$ will have a term in the mgf of $p_1e^t$.

A probability of $p_2$ at $X=2$ will have a term in the mgf of $p_2e^{2t}$.

A probabiity of $p_i$ at $X=x_i$ will have a term in the mgf of $p_ie^{x_it}$.

This all follows directly from the definition of the mgf.

The overall mgf will be a sum of such terms.

Consequently, if you see in the mgf a sum of terms of the form $ae^{bt}$ that means that you have probability $a$ at $X=b$. If you can, simply convert all the terms to be in this form.

Once you have it in the form of a sum of terms like that, you can just write down the probability at any specific value.


As you suggest in your question, the moment generating function holds information on the moments of a distribution. Except for notable examples (e.g. Bernoulli random variable) where the first moment also coincides with the probability of success of the trial, to the best of my knowledge don't hold any direct information on the probability mass.

What you are probably interested in is the probability generating function of a distribution. This on the other hand contains information on the probability mass associated to each value of the random variable's spectrum.

For a discrete non-negative random variable $X$ with $x \in \mathbb{N}$ and probability mass function $p$, then the probability generating function of $X$ is defined as

$$G_X(z) = E(z^X) = \sum_{x=0}^\infty p(x)z^x$$

The logic of it is very similar to that of a moment generating function: a clothesline where instead of hanging moments we hang probabilities of points on the spectrum. Taking the appropriate derivative and evaluating the function at 0 we obtain

$$P(X=k) = \frac{d^{(k)}}{dt^k}\frac{G_X(0)}{k!}$$


I'm sorry, I did not read the question thoroughly enough the first time, I'll clarify better off now.

One way you could go by doing this is using the connection between the mgf and the pgf of a discrete random variable. Using your definition of moment generating function and my definition of probability generating function, we can say that

$$M_Y(t) = E(e^{tY}) = G_Y(e^t)$$

and so a way to obtain probabilities directly from the moment generating function could be the following

$$\frac{1}{k!}\frac{d^{(k)}}{dt^k}M_Y(log(t)) \quad \text{evaluated at} \quad t=0$$

In your example, we would have

$$M_Y(log(t)) = \frac{2^{log(t)} + e^{3log(t) + 4k}}{3}$$

If you wanted to find $P(X=1)$, you would take the first derivative and let $t=0$.

  • $\begingroup$ Thank you for your answer. Suppose Y in my example is a discrete random variable that admits values other than positive integers, then is it possible to know Pr(Y=1) without considering probability generating function? $\endgroup$
    – hephaes
    Nov 3, 2018 at 16:03
  • $\begingroup$ I have added some information. This still only works for a non-negative discrete random variable. If you want to expand the support, then one possible way to go would be to you a Fourier inversion, but it's slippery terrain $\endgroup$
    – Easymode44
    Nov 3, 2018 at 16:57
  • $\begingroup$ Thank you. But my prof says this is actually solvable without regards to pgf, I am now thinking about if the moment generating function corresponds to any common probability distribution, but I can’t find one, do you have any idea? $\endgroup$
    – hephaes
    Nov 3, 2018 at 17:08
  • $\begingroup$ Sorry for the mistake I made in the answer (I read it as $e^t$ instead of $2^t$), which creates problems because of and indeterminate form when evaluating the first derivative at $t=0$. The formula however should hold, and you might be able to get out of it by using l'Hopital's rule if the conditions hold. With regards to your mgf, it does not remind me of any that I know. I would not know how to go by solving this problems without resorting to pgfs! $\endgroup$
    – Easymode44
    Nov 3, 2018 at 17:16

First, for a mgf $\DeclareMathOperator{\E}{\mathbb{E}} M_Y(t)=\E e^ {tY}$ we have that $M_Y(0)=1$. Using that you can find $4k=1$. Using as in the other answer the corresponding probability generating function (assuming that the given function is really the mgf of some random variable with support on the nonnegative integers), we could find the point probabilities by a series expansion. I did so in maple, but that gives a series with many negative coefficients, indicating that the assumption about the support is incorrect. An example where this approach works. So we should stay with the mgf.

If the mgf exists, in the sense that it exist for all arguments in some open interval containing zero, then it uniquely characterizes distributions, so that if two random variables have the same mgf (existing in above sense), then they have the same distribution. So, in principle $\mathrm{Pr}(Y=1)$ can be determined. But in practice that can be difficult. First, if $Y$ is continuous, then $\mathrm{Pr}(Y=1) =0$. But I do not know if/how we can determine from the mgf alone if the distribution is continuous ... In most practical (even theoretical) problems, apart from your quiz-type problem, we know beforehand at least the support of the random variable. Without that knowledge even numerical methods will have problems ...


Firstly, it is useful to note that the requirement that $M_Y(0) = 1$ means that we must have $k = \tfrac{1}{4}$. Hence, this constant is fixed, and we can write the characteristic function of $Y$ as:

$$\varphi_Y(t) = \frac{1 + e^{(\ln 2) i t} + e^{3 i t}}{3} = \frac{e^{0 i t} + e^{(\ln 2) i t} + e^{3 i t}}{3}.$$

We can obtain the probability mass function for $Y$ by appropriate inversion of the characteristic function. As a preliminary matter we note that:

$$\frac{1}{2T} \int \limits_{-T}^{+T} e^{i k t} dt = \text{sinc} (kT) \quad \quad \quad \quad \lim_{T \rightarrow \infty} \frac{1}{2T} \int \limits_{-T}^{+T} e^{i k t} dt = \mathbb{I}(k=0).$$

Hence, using a well-known Fourier inversion formula we obtain:

$$\begin{equation} \begin{aligned} p_Y(y) &= \lim_{T \rightarrow \infty} \frac{1}{2T} \int \limits_{-T}^{+T} e^{-ity} \varphi_Y(t) dt \\[6pt] &= \lim_{T \rightarrow \infty} \frac{1}{2T} \int \limits_{-T}^{+T} \frac{e^{-iyt} + e^{(\ln 2 - y) i t} + e^{(3-y) i t}}{3} dt \\[6pt] &= \frac{\mathbb{I}(y = 0)+ \mathbb{I}(y = \ln 2) + \mathbb{I}(y = 3)}{3}. \\[6pt] \end{aligned} \end{equation}$$

Hence, we see that $Y$ is uniformly distributed over the three points $Y=0,\ln 2, 3$. (In fact, as Glen_b points out in his answer, with a bit of familiarity with Fourier transformation, this form can easily be seen from the original characteristic function, without having to do the maths.)


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