I'm trying to apply the following corollary of Lebesgue's Dominated Convergence theorem:
"Suppose the function $h(x,y)$ is continuous at $y_{0}$ for each $x$, and there exists a function $g(x)$ satisfying:
i. $|h(x,y)| \le g(x)$ for all x and y,
ii. $\int_{-\infty}^{\infty} g(x)dx < \infty$
Then,
$\lim_{y \rightarrow y_{0}}\int_{-\infty}^{\infty} h(x,y)dx = \int_{-\infty}^{\infty} \lim_{y \rightarrow y_{0}}h(x,y)dx$"
…to the below function's central moment generating function to find the 2nd, 3rd, & 4th central moments as follows:
$f(x) = e^{-x}$ $\quad$ $x \ge 0$
$C_{x}(t) = E(e^{t(x-\mu)}) = e^{-t\mu}E(e^{tx}) = e^{-t\mu}M_{x}(t)$
Where C and M represent the central mgf and the mgf, respectively.
My result was as follows:
2nd central moment = $\frac{\partial}{\partial t} \int_{0}^{\infty}e^{-t\mu}.e^{tx}.e^{-x} dx \;\lvert_{t=0}$
applying the above corollary yields:
$\int_{0}^{\infty} e^{-x}(\frac{\partial}{\partial t}e^{-t\mu}.e^{tx}) dx \;\lvert_{t=0}$
= $\frac{((t-1)x-1)e^{tx-x-t}}{(t-1)^2}\lvert_{0}^{\infty}$
at t=0, this evaluates to 0
However, I know for a fact that the second central moment is 1
Can someone help me figure out what's going wrong here? I'm guessing that the second requirement displayed above is not satisfied to be able to do the interchange.
Best Answer
To justify the interchange of limit and integral, estimate the difference quotient as follows: Let $|t|<\delta<1$,
$$ \left|\frac{e^{-t(\mu-x)}-1}{t}\right|e^{-x}\leq e^{-x}\sup_{|t|<\delta}|x-\mu|e^{-t(\mu-x)}\leq e^{-(x-\delta|x-\mu|)}|x-\mu| $$ Which is integrable. Now apply the DCT to find $$ \lim_{t\to 0}\int_0^\infty \frac{e^{-t(\mu-x)}-1}{t}e^{-x}\mathrm dx =\int_0^\infty \lim_{t\to 0}\frac{e^{-t(\mu-x)}-1}{t}e^{-x}\mathrm dx\\ =\int_0^\infty (x-\mu)e^{-x}\mathrm dx\\ =1-\mu $$ Which computes the first central moment, as far as I'm aware.