On Moment-generating Function

Question

This is Exercise $38$ from Tao' note: https://terrytao.wordpress.com/2015/10/03/275a-notes-1-integration-and-expectation/comment-page-1/#comment-681309

Let ${X}$ be a real random variable with cumulative distribution function $${F_X(x) := {\bf P}(X \leq x)}$$ Show that

$$\displaystyle {\bf E} e^{tX} = \int_{\bf R} (1-F_X(x)) t e^{tx}\ dx$$

for all ${t > 0}$.

For this problem a PDF is not assumed to exist (except in a distributional or measure theoretic sense), so care would have to be taken if one wished to proceed by using the fact that the derivative of CDF is PDF. Instead, Tao suggests that one can approximate the exponential function by a piecewise constant function and use dominated convergence to justify passing to the limit.

By the change of variables formula, we want $$\int_{\bf R} e^{tx}\ d\mu_X(x) = \int_{\bf R} (1-F_X(x)) t e^{tx}\ dx$$ where $\mu_X$ denotes the distribution of $X$, how should one choose the approximating p.c function?

Answer 1

This should do the trick. $$\begin{align*}E\exp (tX) = \int _0^\infty \mathbb P\{\exp (tX) \geqslant x\}dx &= \int _0^\infty \mathbb P \left\{ X\geqslant \frac{\ln x}{t}\right\}dx \\ &= \int _0^\infty \left(1-F_X\left(\frac{\ln x}{t}\right)\right) dx \\ &=\int _{-\infty}^\infty \left(1-F_X(s)\right)te^{ts}ds,\quad (t>0). \end{align*}$$

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As for first equality, using the standard machine one can show that for nonnegative random variables it holds that $$ EX = \int _0^\infty \mathbb P\{X\geqslant x\}dx. $$ Perhaps this was alluded to with the comment regarding piecewise constant maps, because that's a step in the standard machine: prove the statement for nonnegative simple measurable functions and then use the fact that any nonnegative measurable function is a pointwise monotone limit of simple functions.