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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.

Elie Fraser
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2 Answers2

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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.

Alijah Ahmed
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  • When you took the first derivative, I think you forgot to multiply $ae^{at}$ by $(1-bt^2)$ – Elie Fraser Jun 26 '15 at 19:43
  • 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 $\color{red}{\frac{d}{dt}e^{at}}\times\frac{1}{1-bt^2}+e^{at}\color{blue}{\times\frac{d}{dt}\frac{1}{(1-bt^2)}}=\frac{\color{red}{ae^{at}}}{(1-bt^2)}+e^{at} \color{blue}{\frac{2bt}{(1-bt^2)^2}}$ – Alijah Ahmed Jun 26 '15 at 20:00
  • 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? – Elie Fraser Jun 26 '15 at 20:16
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    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. – Alijah Ahmed Jun 26 '15 at 20:25
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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.

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