Let $X$ and $Y$ be identically distributed independent random variables such that the moment generating function of $X+Y$ is
$M(t) = 0.09 e^{−2t} + 0.24 e^{−t} + 0.34 + 0.24 e^t + 0.09 e^{2t}$.
Calculate $P(X \le 0)$.
My attempt
Let $u=e^t$
$M_{X+Y}(t)=\frac{1}{(10u)^2}(9+24u+34u^2+24u^3+9u^4)$
Now let $v=u+\frac{1}{u}$ and $p(u)=9u^4+24u^3+34u^2+24u+9$
$p(u)=u^2[9(u^2+\frac{1}{u^2})+24(u+\frac{1}{u})+34]=u^2[9(v^2-2)+24v+34]=u^2(9v^2-18+24v+34)=u^2(9v^2+24v+16)=u^2(3v+4)^2=u^2(3u+\frac{3}{u}+4)^2=[u(3u+\frac{3}{u}+4)]^2=(3u^2+4u+3)^2$
Thus $M_{X+Y}(t)=\frac{1}{(10u)^2}(3u^2+4u+3)^2=[\frac{(3u^2+4u+3)}{10u}]^2=(0.3u+0.4+0.3u^{-1})^2=(0.3e^t+0.4+0.3e^{-t})^2$
Because $X$ and $Y$ are independent and identically distributed, the MGF of $X+Y$ equals $(0.3e^t+0.4+0.3e^{-t})^2$, where $0.3e^t+0.4+0.3e^{-t}$ is the MGF common to $X$ and $Y$.
Thus $M_X(t)=0.3e^t+0.4+0.3e^{-t}$, and since $M_X(t)$ uniquely determines the distribution of $X$, it follows that $X$ is a discrete random variable whereby $X=-1, X=0$, and $X=1$ with respective probabilities $0.3,0.4$, and $0.3$.
$\therefore P(X \le 0)=0.3+0.4=0.7$.
When is it appropriate to use the change of variable technique (i.e. $v=u+\frac{1}{u}$) to factor polynomials and in what situations would it not work?
Best Answer
The method works for Quasi-palindromic quartic functions as $$f(x)=a_0\cdot x^4+a_1\cdot x^3+a_2\cdot x^2+a_1\cdot m\cdot x+a_0 \cdot m^2$$
In you case the parameters are $m=1, a_0=9, a_1=24$ and $a_2=34$. So you have a (real) palindromic function. For reference see here.
If you have not a palindromic quartic function, then you can multiply out $g(x)=(ax^2+bx+c)^2$ and then you compare the coefficients.