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.

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

1 Answer 1


$X$ 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}

  • $\begingroup$ 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. $\endgroup$
    – Onix
    Jan 24, 2016 at 12:02
  • $\begingroup$ @Abomm Yes, you're correct. $\endgroup$
    – Mick A
    Jan 24, 2016 at 12:11

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