Problem:Let $X$ and $Y$ be identically distributed independent random variables such that the moment generating function of $X +Y$ is $$M(t) = 0.09 e^{−2t} + 0.24 e^{−t} + 0.34 + 0.24 e^t + 0.09 e^{2t}$$ for $−\infty < t < \infty$. Calculate $P[X \le 0]$.

Answer: Because $X$ and $Y$ are independent and identically distributed, the moment generating function of $X+ Y$ equals $K^2(t)$, where $K(t)$ is the moment generating function common to $X$ and $Y$. Thus, $$K(t) = 0.30e^{-t} + 0.40 + 0.30e^t$$

This is the moment generating function of a discrete random variable that assumes the values $-1$, $0$, and $1$ with respective probabilities $0.30$, $0.40$, and $0.30$. The value we seek is thus $0.70$. My question is how to factor the moment generating function of $X+Y$ to $0.30e^-t + 0.40 + 0.30e^t$, is there a general formula to use?

I'm also sorry for not using MathJax I'm really having trouble with it.

  • $\begingroup$ I fixed the math formatting. Check if it's ok. And you should look into the edited question (press "edit") to learn by example how it's done, it's not difficult. $\endgroup$ – leonbloy Aug 11 '16 at 22:57
  • $\begingroup$ Let $u = e^t$. Start by noting that $0.09u^2 + 0.09 u^{-2} = \left(0.03 ( u + u^{-1})\right)^2 - 0.18$. Next set $u + u^{-1} = v$. You now have the quadratic $(0.03 v)^2 + 0.24 v + 0.16$, which is easy to factorise. BTW - the fact that $(x + x^{-1})^2 = x^2 + x^{-2} + 2$ is in general very useful, and should immediately enter your mind when trying to factorise expressions which have terms like $ax^2 + ax^{-2}$. $\endgroup$ – stochasticboy321 Aug 11 '16 at 23:01
  • $\begingroup$ As an aside, in this case it's possible to construct $K$ merely by inspection. You know that $X$ and $Y$ are iid, and since their sum is in $\{-2, -1, 0,1,2\}$, they must take values in $\{-1,0,1\}$. Now, the sum can only equal $-2$ if they're both $-1$, so you immediately know that $P(X = -1) = \sqrt{0.09}$, and similarly for $P(X=1)$. Normalisation will then give you $P(X=0)$. $\endgroup$ – stochasticboy321 Aug 11 '16 at 23:05
  • $\begingroup$ Thanks, I would never thought of this on the test $\endgroup$ – Theo Robinson Aug 12 '16 at 15:26

The factorization may be easier to detect if we let $z = e^t$ and write $$\begin{align*} M_{X+Y}(t) &= \frac{9}{100} z^{-2} + \frac{24}{100} z^{-1} + \frac{34}{100} + \frac{24}{100} z + \frac{9}{100} z^2 \\ &= \frac{1}{(10z)^2} \left( 9 + 24 z + 34 z^2 + 24 z^3 + 9 z^4 \right) \\ &= \frac{1}{(10z)^2} (3 + 4z + 3z^2)^2, \end{align*}$$ keeping in mind that you are looking for a perfect square. Alternatively, you can posit a form of the common MGF to $X$ and $Y$, observing from the powers of $e^t$ that it needs to be $$M_X(t) = M_Y(t) = az^{-1} + b + cz$$ for suitable constants $a, b, c$, then equating coefficients of $M_X (t)^2$.

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    $\begingroup$ To factor every polynomial $$p(z)=a+bz+cz^2+bz^3+az^4$$ use the change of variable $$u=z+\frac1z$$ and note that $$p(z)=z^2\,\left(a(z^2+1/z^2)+b(z+1/z)+c\right)=z^2\,\left(a(u^2-2)+bu+c\right)$$ then one can factor the quadratics in $u$, leading to $$p(z)=az^2(u-\alpha)(u-\beta)=a(z^2-\alpha z+1)(z^2-\beta z+1)$$ and now one must simply factor the quadratics $z^2-\alpha z+1$ and $z^2-\beta z+1$. If $$(a,b,c)=(9,24,34)$$ then $$a(u^2-2)+bu+c=9u^2+24u+16=(3u+4)^2$$ hence, indeed, as stated above, $$p(z)=(3z^2+4z+3)^2$$ $\endgroup$ – Did Aug 27 '16 at 10:04

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