Consider a random variable X with the following moment generating function:


(a) Find the expected value of $3 - X$.

(b) Find variance of $3 - X$.

What I have reached till now is that is (most probably?) geometric distribution which means that the probability is 0.2, and that if we want to get the expected value of $3 - X$ then it would be equal to $E(3-X) = E(3) - E(X)$

What I think: E(X) = 1/0.2 But I have no idea where to go from there.


a) $E(3-X) = E(3) - E(2Y) = E(3) - 2 * E(Y) = 3 - 2 * 1/0.2 = -7$ (I also used differentiation rule to double check, but differentiation is way longer)


I tried to use a similar logic to solve (b), since the variance = $p/(1-p)^2$ I thought I could equate $Var(X) = Var(2Y) = 2 * (0.8/0.2^2) = 40$

Not sure if this method is right or not, though..


Answer of B: Okay, so, $Var(X) = Var(2Y) = 2^2 Var(Y) = 4 * (0.8/0.2^2) = 80$

Thank you all! :)

  • $\begingroup$ Hint: The expected value of a constant is that constant. $\endgroup$
    – Mosquite
    Oct 18, 2016 at 18:26
  • $\begingroup$ Your edit is right. $\endgroup$ Oct 18, 2016 at 18:55
  • $\begingroup$ You can proof your result of the variance by using differentiations. $\endgroup$ Oct 18, 2016 at 19:07
  • $\begingroup$ The variance of a translation of a random variable is the same as that for the random variable. So $Var(3-X)=Var(X)$. $\endgroup$
    – Paul
    Oct 18, 2016 at 19:10
  • $\begingroup$ Yeah. $Var(X)=80$ is right. $\endgroup$ Oct 18, 2016 at 19:16

3 Answers 3


The property of moment-generating functions that you should recall for this problem is $$\left[\frac{\partial^k M_X(t)}{\partial t^k}\right]_{t=0} = M^{(k)}_X(0) = \operatorname{E}[X^k];$$ that is to say, the $k^{\rm th}$ derivative of the MGF at $t = 0$ is the $k^{\rm th}$ raw moment of $X$, whenever such a moment is defined. So, for $k = 1$, we observe that the first derivative of $M_X$ at $t = 0$ gives the expectation; for $k = 2$, we get the expectation of $X^2$. Then to obtain the variance, you would calculate $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$

To understand where this relationship comes from, recall that the MGF is defined as $$M_X(t) = \operatorname{E}[e^{tX}] = \operatorname{E}\left[\sum_{k=0}^\infty \frac{(tX)^k}{k!}\right] = \sum_{k=0}^\infty \frac{\operatorname{E}[X^k]}{k!} t^k.$$ But by Taylor's theorem, $$M_X(t) = \sum_{k=0}^\infty \frac{M_X^{(k)}(0)}{k!} t^k,$$ thus $\operatorname{E}[X^k] = M_X^{(k)}(0)$ as claimed (again, whenever the moments are defined).


If $Y$ is geometric with success probability $0.2$, then $m_Y(t)=\frac{0.2 e^t}{1-0.8 e^t}$. You can show that then $X=2Y$ has your given MGF $m_X(t)=m_Y(2t)=m_{2Y}(t)$.


Hint: Differentiate $m(t)=\large{\frac{0.2\cdot e^{2\cdot t}}{1-0.8\cdot e^{2\cdot t}}}$ w.r.t. $t$. For this purpose apply the quotient rule.

Then set $t=0$ to get the expected value. In short form


And the variance is $Var(X)=m''(0)-[m'(0)]^2$ and $Var(3-X)=Var(X)$

  • $\begingroup$ No. We have $\operatorname{Var}[X] = m''(0)$ if and only if $\operatorname{E}[X] = 0$. $\endgroup$
    – heropup
    Oct 18, 2016 at 18:42
  • $\begingroup$ Thanks for the comment, I have made an edit. $\endgroup$ Oct 18, 2016 at 18:44

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