# Is there any relation between the moment generating function (mgf) of a discrete probability distribution and probability at X=0?

I have encountered this question in my test today, can't solve it but what is the possible solution.

the moment generating function of an integer valued random variable $X$ is given by $$M_X (t)=\frac{1}{10}\left(2+e^t+4e^{2t}+3e^{3t}\right)e^{-t}$$ Then $P(2X+5<7)=$

1. $\frac{3}{10}$
2. $\frac{7}{10}$
3. $\frac{4}{10}$
4. $1$

If I manipulate the parenthesis we need to find $P(X<1)=P(X=0)$ I could not understand the context please provide some hint how to use all this information. If my logic is not correct please adjust the title accordingly.

• $P(X<1)=P(X=0)$ and not $P(X<1)=1-P(X=0)$ as you have it, right? Jan 24, 2016 at 11:00
• Thanks for notifying.
– Onix
Jan 24, 2016 at 11:03

$X$ is an integer so you shouldn't assume $X\lt 1$ means $X=0$ because $X$ could be negative.

If you re-arrange the mgf slightly, the pmf will become obvious:

\begin{align} M_X(t) &= \dfrac{2}{10}e^{-t} + \dfrac{1}{10}e^{-0t} + \dfrac{4}{10}e^{t} + \dfrac{3}{10}e^{2t} \\ \end{align}

So we can see that:

\begin{align} P(X=-1) &= \dfrac{2}{10} \\ P(X=0) &= \dfrac{1}{10} \\ P(X=1) &= \dfrac{4}{10} \\ P(X=2) &= \dfrac{3}{10} \\ & \\ \text{giving }\quad P(X\lt 1) &= \dfrac{3}{10}. \end{align}

• Really awesome, and we can take only upto X=-1 integral value because the mgf has only $e^{-t}$ term, am I correct and thank you so much for providing the answer. Really helpful and taught me sometimes i don't really look things patiently.
– Onix
Jan 24, 2016 at 12:02
• @Abomm Yes, you're correct. Jan 24, 2016 at 12:11