I'm working through Schaum's Outline of Probability, Random Variables, and Random Processes, and am stuck on a question about moment-generating functions. If anyone has the 2nd edition, it is question 4.60, part (b).

The question gives the following initial information: $$E[X^k] = 0.8$$ for k = 1, 2, ...

The moment-generating function for this is the following: $$0.2 + 0.8\sum_{k=0}^\infty\frac{t^k}{k!} = 0.2 + 0.8e^t$$

The question is asking to find $P(X=0)$ and $P(X=1)$. The answers are given, $P(X=0)=0.2$ and $P(X=1)=0.8$, but I'm not seeing how the book arrived at these answers.

Using the definition of moment-generating functions, I see that the following equation is utilized: $$\sum_{i}e^tx_i*p_X(x_i) = 0.2 + 0.8\sum_{k=0}^\infty\frac{t^k}{k!} = 0.2 + 0.8e^t $$

But I'm not seeing how the $p_X(x_i)$ is extracted from that equation.

Any help is greatly appreciated. Thanks.


2 Answers 2


Note that the distribution of the random variable $Y$ which is $0$ with probability $0.2$ and $1$ with probability $0.8$ has the same mgf: to check, compute. Now use uniqueness.

Like in the case of the Laplace Transform, we often recognize an mgf as being a familiar one, and thereby identify a distribution.

Detail: Let $Y$ be a Bernoulli random variable which is $0$ with probability $a$, and $1$ with probability $b$. Then $$E(e^{tY})=a e^{(0)t}+be^{(1)(t)}=a+be^t.$$ Conversely, by the uniqueness theorem, a random variable whose distribution has mgf $\,a+be^t$ is a Bernoulli random variable, taking on the value $0$ with probability $a$, and $1$ with probability $b$.

We have found that our random variable $X$ has distribution with mgf $\,0.2+0.8e^t$. Therefore $X$ must be a Bernoulli random variable, with $\Pr(X=0)=0.2$ and $\Pr(X=1)=0.8$.

  • $\begingroup$ Andre, thanks for your response, but I don't feel it answered my question. I'm not seeing how the answers were obtained. Assuming you didn't know that $P(X=0) = 0.2$ and $P(X=1) = 0.8$, how would you go about solving for those two probabilities? $\endgroup$
    – konfushus
    Sep 21, 2012 at 3:30
  • $\begingroup$ You would recognize that $a+be^t$ is the mgf of a distribution which takes on value $0$ with probability $a$ and $1$ with probability $b$. Then you read off $a$ and $b$ from the mgf you computed. $\endgroup$ Sep 21, 2012 at 3:33
  • $\begingroup$ Sorry Andre, I'm still not following. I want to see the answer mathematically, not through just "looking at it". If looking at $P(X=0)$ first, let me plug that into the equation above. That leaves me with, $$e^{t(0)}p_X(0) = 0.2 + 0.8e^t$$ In order for $p_X(0) = 0.2$, the $0.8e^t$ must disappear. That's what I'm not seeing. Where does it go? Thanks again for your help. $\endgroup$
    – konfushus
    Sep 22, 2012 at 22:35
  • $\begingroup$ @konfushus: I have added to the post, but really only repeating in symbols what was written before. $\endgroup$ Sep 22, 2012 at 23:25
  • $\begingroup$ Andre, thanks for the added detail. I see now that the question wanted me to recognize that the random variable in question is, in fact, a Bernoulli random variable, thus having a sample space of just 0 and 1. If plugging 0 and 1 into the LHS and comparing like co-efficients, I see how the solution is obtained. Thanks again for your help. $\endgroup$
    – konfushus
    Sep 23, 2012 at 14:34

It is easy, we know that

$$ M_X(log_e(t))= G_X(t) $$

where $M_X(\bullet)$ denotes Moment generating function of X and $G_X(\bullet)$ represents Probability generating function of X, So we have to generally replace $t$ by $log_e(t)$ by doing that with the MGF you have given we will get

$$M_X(log_e(t))=0.2 + 0.8e^{log_et}$$

$$G_X(t)=0.2 +0.8t$$

as we know that $$G_X(t)=p_0+p_1t+p_2t^2\ldots\ldots\ldots\ldots$$


$$p_0 = P(X=0)$$

$$p_1= P(X=1)$$




as in our Generating function we can see that $$p_0= 0.2 =P(X=0)$$

$$p_1= 0.8 =P(X=1)$$



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