Esscher Tilted Expectations as Log-MGF Derivatives
Summary
The document examines why integrating a variable against an Esscher-transformed distribution is equivalent to weighting the original distribution by an exponential factor and dividing by the moment-generating function. Under this change of measure, the expectation becomes the ratio of the derivative of the moment-generating function to the function itself, which is the derivative of its logarithm at the tilt parameter.
The answer motivates the measure change by relating distribution-function integration to density integration, then differentiates the weighted cumulative integral. That explanation explicitly uses a density and invokes differentiation under the integral in the absolutely continuous case. The measure identity also holds more generally through the definition of the tilted measure, but the density-based derivation shown does not itself explain discrete or singular distributions. Existence of the relevant moment-generating function and derivative is also required.
Key ideas
- The Esscher distribution weights the original measure by an exponential factor and normalizes it by the moment-generating function.
- Its expected value is the ratio of the moment-generating function’s derivative to the function itself.
- That ratio equals the derivative of the log moment-generating function at the tilt parameter.
- The answer derives the measure transformation through densities and differentiation of a cumulative integral.
- The density argument needs adaptation for distributions without a density.
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# Esscher Premium: Integral Transform Proof
# Esscher Premium: Integral Transform Proof
I have some difficulty understanding the following proof and I hope someone can help me with that.
Claim: I want to show that
$E_\alpha(S)=\frac{d}{dr} \log M_S(r)|_{r=\alpha} $, where $M_S(r)=E(\exp(rS))$ is the moment generating function of $S\sim F$ and $E_\alpha(S)=\int_{\mathbb{R}} s dF_\alpha(s)$ with $F_\alpha(s):=\frac{1}{M_S(\alpha)}\int_{-\infty}^s e^{\alpha x} dF(x)$.
Proof The proof goes a follows: $E_{\alpha}(S)=\int_{\mathbb{R}} s dF_\alpha(s)=^{(*)}\frac{1}{M_S(\alpha)}\int_{\mathbb{R}}s e^{\alpha s} dF(s)=\frac{M_S'(\alpha)}{M_S(\alpha)}=\frac{d}{dr} \log(M_S(r))|_{r=\alpha}$.
What I don't understand ist the second equality $(*)$. Why and how can this transformation from $dF_\alpha(s)$ to $dF(s)$ be done like that?
I appreciate any hints! :-) Thanks in advance!
## Answer by Pleb (score 2, accepted)
https://quant.stackexchange.com/a/60426
Okay, I've tried to come up with a solution. We know that $\int x \: dF(x)$ is a generalization of $\int x f(x) \: dx$, since:
$$\frac{dF(x)}{dx}=f(x) \iff dF(x)=f(x)\: dx$$ for $f$ being the pdf and $F$ the CDF. Using this result on $F_{\alpha}(s)$, we get:
\begin{align} \frac{dF_{\alpha}(s)}{ds}&= \frac{1}{M_{S}(\alpha)} \cdot \frac{d}{ds}\left(\int_{-\infty}^s e^{\alpha x} dF(x)\right)\\ &=\frac{1}{M_{S}(\alpha)} \cdot \frac{d}{ds}\left(\int_{-\infty}^s e^{\alpha x} f(x) \: dx\right)\\ &=\frac{1}{M_{S}(\alpha)} \cdot e^{\alpha s} \cdot f(s),\\ \end{align} where I've used the first fundamental theorem of calculus for Lebesgue integrals (the Lebesgue differentiation theorem) in the last equation (as stated in a comment below). Therefore we observe that: $$dF_{\alpha}(s) = \frac{1}{M_{S}(\alpha)} \cdot \left( e^{\alpha s} \cdot f(s) \cdot ds \right).$$ Now, inserting this into the integral specified in your above proof, we get: \begin{align} \int_{\mathbb{R}} s \: dF_{\alpha}(s) &= \frac{1}{M_{S}(\alpha)} \int_{\mathbb{R}} s \cdot e^{\alpha s} \cdot f(s) ds\\ &= \frac{1}{M_{S}(\alpha)} \int_{\mathbb{R}} s \cdot e^{\alpha s}\: dF(s) \end{align} Giving you the second equality.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.