Differentiating a Moment-Generating Function Under the Expectation
Summary
The document explains why differentiating a moment-generating function requires justification for moving a limit through an expectation. A density-based calculation can suggest the result, but continuity of the integrand in the parameter alone does not establish that the derivative and integral may be interchanged. The cited proof instead writes a difference quotient, applies the mean value theorem pointwise, and uses dominated convergence to pass its limit through the expectation.
The discussion also notes how to extend the argument from nonnegative random variables to integrable variables by splitting them into positive and negative parts. The example is a proof technique, not a trading strategy or empirical result. Its conditions matter: the exponential moments must provide an integrable dominating bound near the parameter of interest, and integrability of X alone does not guarantee the moment-generating function is finite or differentiable at every t.
Key ideas
- Differentiating under an expectation requires conditions that justify exchanging a limit and expectation.
- A density is not necessary because the argument can be stated directly for random variables.
- The mean value theorem bounds the difference quotient, while dominated convergence justifies taking its limit inside the expectation.
- Positive and negative parts extend the reasoning beyond nonnegative random variables.
- The required exponential integrability depends on the parameter and cannot be inferred from integrability of X alone.
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# Proving Derivative Property of Moment-Generating Function
# Proving Derivative Property of Moment-Generating Function
In Shreve II, exercise 1.8, he walks the reader through proving the derivative of a moment-generating function $\phi$ is equal to the expectation $\mathrm{E}[Xe^{tX}]$; i.e., $$ \phi^\prime(t) = \mathrm{E}[Xe^{tX}]. $$ He does this first assuming $X$ is nonnegative and asks the reader to use the dominated congergence theorem along with the mean value theorem, then asks the reader to prove it for an integrable $X$ by considering the positive and negative parts.
My question is, is all this machinery necessary for this exercise? Assuming $X$ has a density function $f$, I would prove this result by $$ \phi^\prime(t) = \frac{d}{dt}\int_{0}^\infty e^{tx}f(x) dx = \int_{0}^\infty \frac{d}{dt}e^{tx}f(x) dx = \int_{0}^\infty xe^{tx}f(x) dx = \mathrm{E}[Xe^{tX}], $$ where I guess I would have to justify moving the derivative inside the integral (is it enough that $e^{tx}$ is continuous in $t$?)
So, is this more rigorous proof in Shreve just to show it holds if $X$ doesn't have a density? Or perhaps just to practice using the dominated convergence theorem? I wish authors would state the spirit behind their exercises...
## Answer by user16891 (score 2)
https://quant.stackexchange.com/a/18975
As Sherev has said, first let $\varphi(t)=E\left[e^{tX}\right]$ then $$\varphi '(t)=\underset{s\to t}{\mathop{\lim }}\,\frac{\varphi (t)-\varphi (s)}{t-s}=\underset{s\to t}{\mathop{\lim }}\,\frac{E[{{e}^{tX}}]-E[{{e}^{sX}}]}{t-s}=\underset{s\to t}{\mathop{\lim }}\,E\left[ \frac{{{e}^{tX}}-{{e}^{sX}}}{t-s} \right]$$ Sherev continue ,we can choose a sequence of numbers $\{s_n\}_{n=1}^{\infty}$ that converges to $t$ and compute $$\underset{s_n\to t}{\mathop{\lim }}\,E\left[ \frac{{{e}^{tX}}-{{e}^{s_nX}}}{t-s_n} \right]$$ where now we are taking a limit of the expectations of the sequence of random variables. $$Y_n=\frac{{{e}^{tX}}-{{e}^{s_nX}}}{t-s_n}$$ he fixes $\omega\in \Omega$ and by application of Mean Value Theorem concludes $$e^{tX(\omega)}-e^{s_nX(\omega)}=(t-s_n)X(\omega)e^{\theta(\omega )X(\omega)}$$ where $\theta(\omega)$ is a number depending on $\omega$ such that $s_n\leq\theta(\omega)\leq t$.
$$|Y_n|=\left|\frac{e^{tX(\omega)}-e^{s_nX(\omega)}}{t-s_n}\right|\leq X(\omega) e^{\theta(\omega )X(\omega)}\leq X(\omega) e^{2t\,X(\omega)}$$ So by the Dominated Convergence Theorem we have $$\varphi'(t)=\underset{n\to \infty }{\mathop{\lim }}\,E[{{Y}_{n}}]=E[\underset{n\to \infty }{\mathop{\lim }}\,{{Y}_{n}}]=E[X{{e}^{tX}}]$$
## Answer by LazyCat (score 0)
https://quant.stackexchange.com/a/21818
The whole machinery is needed to interchange the integral sign and the derivative.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.