The Moment Generating Function of a standard Gaussian distribution is $\exp(t^2/2)$. Let $X$ be a Gaussian r.v., $a>0$ and define the (outer-)truncated variable $Y=X\mathbb{I}(|X|\ge a)$. What is the MGF $E[\exp(tY)]$ of $Y$? It should be upper-bounded by $\exp(t^2/2)$, but should not be the same, since the moments of the two distributions are strictly lower. Even an upper bound on the function that is smaller than the gaussian case would be interesting.
As a variant, I would be interested in comparing the MGF of the folded normal $|X|$ against that of $|Y|$.
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$\begingroup$We can write the MGF$$ \phi(t) = \Big\langle \exp\big[t X I(|X|>a) \big]\Big\rangle_X. $$Because $I(|X|>a)$ is either $1$ or $0$ we have$$ \phi(t) = \Big\langle 1- \big(1-e^{t X}\big)I(|X|>a)\Big\rangle_X. $$or$$ \phi(t) = 1 - \Big\langle I(|X|>a) \Big\rangle + \Big\langle e^{tX}I(|X|>a) \Big\rangle.$$Writing out the integrals,$$ \phi(t) = 1 -\int_{-\infty}^{-a} \frac{e^{-x^2/2}}{\sqrt{2\pi}}dx - \int_a^\infty \frac{e^{-x^2/2}}{\sqrt{2\pi}}dx+ \int_{-\infty}^{-a} \frac{e^{tx-x^2/2}}{\sqrt{2\pi}}dx +\int_a^\infty \frac{e^{tx-x^2/2}}{\sqrt{2\pi}}dx.$$Evaluating these gives the desired result:$$ \phi(t) = 1+\frac{1}{2} \left(e^{\frac{t^2}{2}} \text{erfc}\left(\frac{a-t}{\sqrt{2}}\right)-\text{erfc}\left(\frac{a}{\sqrt{2}}\right)\right)+\frac{1}{2} \left(e^{\frac{t^2}{2}} \text{erfc}\left(\frac{a+t}{\sqrt{2}}\right)-\text{erfc}\left(\frac{a}{\sqrt{2}}\right)\right). $$This reduces to the standard result as $a \rightarrow 0$:$$ \phi(t) = e^{t^2/2}.$$You can compare with the folded normal MGF from wikipedia:
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