Probability: Symmetric transform implies symmetric PDF? I am having trouble solving the following problem:
Let $X$ be a continuous random variable. Let $M(s)$ be the transform of $X$, i.e. $M(s) = E[e^{sX}]$. Suppose $M$ satisfies $M(s) = M(-s)$ for any $s$ in the domain of $M.$ Let $f(x)$ be the PDF of $X.$ Show that $f(t) = f(-t)$ for any $t$ in the domain of $f$.
So far I have the following:
$$\ M(s) = E[e^{sX}] = \int_{-\infty}^\infty e^{sx}f(x)dx\\
M(-s) = E[e^{-sX}] = \int_{-\infty}^\infty e^{-sx}f(x)dx\\
M(s) = M(-s) = \int_{-\infty}^\infty e^{sx}f(x)\,dx= \int_{-\infty}^\infty e^{-sx} f(x)\,dx\\
M'(s) = \int_{-\infty}^\infty xe^{sx}f(x)\,dx= \int_{-\infty}^\infty -xe^{-sx} f(x)\,dx\\
M'(0) = \int_{-\infty}^\infty xf(x)\,dx = \int_{-\infty}^\infty -xf(x)\,dx = \int_{-\infty}^\infty xf(-x)\,dx\\
\int_{-\infty}^\infty xf(x)\,dx\ -\int_{-\infty}^\infty xf(-x)\,dx = \int_{-\infty}^\infty x(f(x) \ - f(-x))\,dx = 0
$$
I'm not sure how to proceed from here.
Thanks in advance for your help!
 A: The hypotheses imply that the mgf of $X$ is equal to the mgf of $-X$. In the good case the mgf is finite in neighborhood of $0$, the distribution of $X$ is equal to that of $-X$.  Then all the odd moments of $X$ vanish, and if $X$ has a density function $f(x)$, it too is symmetric: $f(x) = f(-x)$ for almost all $x$.  (The hypothesis that $X$ is continuous does not imply that $X$ has a density function, but the cdf $F(x)$ of $X$ is symmetric in the sense that $F(x)=1-F(-x)$ for all $x$.)  
It might happen that $M(s)<\infty$ only when $s=0$, in which case these deductions do not hold. Call this the bad case.
All of this rests on the fact that in the good case, the moment generating function of a random variable determines its probability distribution uniquely.  That is, packed into the function $M_X(s)$ is complete information of all values $P(X\in A)$, as if in a hologram.  The usual way to see this is to use complex analysis methods to see that $M_X(s)$ is analytic in a vertical strip in the complex plane, and that when restricted to the imaginary axis, $M_X(it)=Ee^{itX}$ is the characteristic function of $X$, which (as is explained in the wikipedia article) determines the probability distribution of $X$.
This is not unrelated to the fact that two functions with the same bilateral Laplace transform must agree.  There is an inversion formula that expresses $P(X\in[a,b])$ in terms of contour integrals of $M_X$.
