1
$\begingroup$

Suppose that the random variable X has moment generating function M(t)= $$e^{at} \over 1-bt^2$$ for -1< t <1. It is found that the mean and variance of X are 3 and 2 respectively. Find a+b.

We have E[X]=3 and Var[X]=2. I know that M'(0)=E[X]. I solved the equation and found that a=3, but I got stuck here. I know that M''(0) should be E[$X^2$] and then I should use Var[X]=E[$X^2$]-$(E[X])^2$ somehow, but I found that M''(0)=0, which didn't make any sense. Where did I go wrong?

Edit:According to the textbook, the answer is 4, if that helps.

$\endgroup$
3
$\begingroup$

Differentiating the MGF once results (using the product rule), and setting $t$ to $0$ results in:- $$\frac{d}{dt}M(0)=\left.\frac{ae^{at}}{(1-bt^2)}+\frac{2bte^{at}}{(1-bt^2)^2}\right|_{t=0}=a=E\{X\}=3$$ Differentiating twice, using the product rule again to each of the two terms of the first derivative, leads (after some simplification) to:- $$\begin{align}\frac{d^2}{dt^2}M(0)&=\left.\frac{a^2e^{at}}{(1-bt^2)}+\frac{4abte^{at}}{(1-bt^2)^2}+\frac{2be^{at}}{(1-bt^2)^2}+\frac{8b^2t^2e^{at}}{(1-bt^2)^3}\right|_{t=0}\\&=a^2+2b=9+2b\\&=E\{X^2\}\\&=Var\{X\}+(E\{X\})^2\\&=11\\\Rightarrow b&=1\end{align}$$ Thus $a+b=4$, as given in the answer sheet.

$\endgroup$
4
  • $\begingroup$ When you took the first derivative, I think you forgot to multiply $ae^{at}$ by $(1-bt^2)$ $\endgroup$ – Elie Fraser Jun 26 '15 at 19:43
  • $\begingroup$ For the first derivative, I expressed $\frac{e^{at}}{(1-bt^2)}$ as $\color{red}{e^{at}}\times\color{blue}{\frac{1}{(1-bt^2)}}$. The derivative will be $\color{red}{\frac{d}{dt}e^{at}}\times\frac{1}{1-bt^2}+e^{at}\color{blue}{\times\frac{d}{dt}\frac{1}{(1-bt^2)}}=\frac{\color{red}{ae^{at}}}{(1-bt^2)}+e^{at} \color{blue}{\frac{2bt}{(1-bt^2)^2}}$ $\endgroup$ – Alijah Ahmed Jun 26 '15 at 20:00
  • $\begingroup$ Oh alright I see what you did there! I tried using the quotient role instead of splitting the fraction, maybe that's why my method didn't work? $\endgroup$ – Elie Fraser Jun 26 '15 at 20:16
  • 2
    $\begingroup$ You could have used the quotient rule, and I reckoned it would have worked too - however, it results in very complicated expressions, where you are more likely to make a mistake. So, I went for the simpler option of splitting the fraction, and using the product rule. $\endgroup$ – Alijah Ahmed Jun 26 '15 at 20:25
1
$\begingroup$

Or you can do it by:

Transforming $M_X(t) = \frac{e^{at}}{1+bt^2}$ to $M_{X-a}(t) = \frac{1}{a+bt^2}$.

Then, $M'_{X-a}(0)=0$ and $M''_{X-a}(0) = 2b$.

$\implies$ $E(X-a) = 0$ and $E(X-a)^2 = b$

This way you can do it a lil bit faster.

$\endgroup$
3
  • $\begingroup$ Wait, I don't get how you transformed $M_X(t) = \frac{e^{at}}{1+bt^2}$ to $M_{X-a}(t) = \frac{1}{a+bt^2}$ $\endgroup$ – Elie Fraser Jun 26 '15 at 19:51
  • $\begingroup$ $M_X(t) = E(e^{tX})$ and $M_X(t-a) = E(e^{t(X-a)})$. In your problem there was an $e^{at}$ term. Understood?? $\endgroup$ – Croma14 Jun 26 '15 at 20:15
  • $\begingroup$ I took the term $e^{at}$ to the left of the equation and got $M_{X-a}(t)$ $\endgroup$ – Croma14 Jun 26 '15 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.