I have a random variable $X$ with moment generating function:

$$m_X(t) = \frac{2}{9} + \frac{e^{-t}}{9} + \frac{e^{-2t}}{9} + \frac{2e^{t}}{9} + \frac{e^{2t}}{3}.$$

I want to find the probability $\mathbb{P}(X \geqslant 0)$.

For me to calculate the probability, I need to identify the distribution of $X$ using uniqueness theorem. However, I am unable to do so. I tried to manipulate the expression in many ways so as to get it into a form where I could express this as product of multiple moment generating functions, but I failed. The MGF resembles the MGF of binomial, but I cannot pinpoint on how to approach this.

  • 2
    $\begingroup$ 1. If you had probabilities of $p$ and $q=1-p$ at 0 and 1, what would the MGF look like? 2. If you had probabilities of $p_1$, $p_2$ and $p_3 = 1-p_1-p_2$ at -2, 0, and 1 respectively, what would the MGF look like? 3. What if you replaced -2 and 1 with $a$ and $b$? 4. Can you see where this is going? $\endgroup$
    – Glen_b
    Commented Feb 21 at 2:04

3 Answers 3


The MGF of a random variable $X$ is defined as $m_X(t) = \mathbb{E}[e^{tX}]$, where $\mathbb{E}$ denotes the expected value. The MGF can be used to determine all the moments of the distribution (like the mean and variance) and, in some cases, to identify the distribution itself.

The given MGF looks like a sum of exponentials with coefficients that could represent probabilities of $X$ taking on specific integer values. This is because $e^{tx}$where $x$ is an integer, would represent the MGF of a discrete distribution at specific points.

To find $\mathbb{P}(X \geqslant 0)$, we need to identify the probabilities of $X$ taking on values of 0 or greater from the given MGF. However, directly extracting these probabilities from the MGF is not straightforward without recognizing a pattern or the form of a known distribution.

The terms in the MGF suggest a distribution that is a mixture of point masses at -2, -1, 0, 1, and 2, with their respective probabilities given by the coefficients. Specifically, we have:

  • The probability of $X = -2$ is $\frac{1}{9}$,
  • The probability of $X = -1$ is $\frac{1}{9}$,
  • The probability of $X = 0$ is $\frac{2}{9}$,
  • The probability of $X = 1$ is $\frac{2}{9}$,
  • The probability of $X = 2$ is $\frac{1}{3}$.

To find $\mathbb{P}(X \geqslant 0)$, we sum the probabilities of $X$ taking on values of 0 or greater:

$$\mathbb{P}(X \geqslant 0) = \mathbb{P}(X = 0) + \mathbb{P}(X = 1) + \mathbb{P}(X = 2).$$


One approach is to say that if the moment generating function is $$\mathbb{E}\left[e^{tX}\right] = \frac{2}{9} + \frac{e^{-t}}{9} + \frac{e^{-2t}}{9} + \frac{2e^{t}}{9} + \frac{e^{2t}}{3}$$

then, since $\mathbb{E}\left[e^{tX}\right]= \mathbb{E}\left[(e^t)^X\right]$, using $s=e^t>0$ gives $$\mathbb{E}\left[s^X\right] = \frac{2}{9}s^0 + \frac{1}{9}s^{-1} + \frac{1}{9}s^{-2} + \frac{2}{9}s^1 + \frac{1}{3}s^2.$$

This is the probability generating function of a discrete random variable $X$

and $\mathbb P(X=n)$ is then the coefficient of $s^n$ in the probability generating function.


Hint: For a discrete random variable with values in $\mathscr{X}$ the moment generating function is:

$$m_X(t) \equiv \mathbb{E}(e^{tX}) = \sum_{x \in \mathscr{X}} e^{tx} \cdot \mathbb{P}(X=x).$$

Does that form for the expected value aid you in figuring out the relationship between the MGF and the probabilities for the random variable?

  • $\begingroup$ Understood. We do not need the distribution to find the answer here $\endgroup$ Commented Feb 21 at 3:40
  • 2
    $\begingroup$ My hint in comments was intended to get you to recognise this equality, or even derive it if needed. $\endgroup$
    – Glen_b
    Commented Feb 21 at 4:21

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