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.
Best Answer
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 $M_X (t)^2$.