Let $M_X(t)$ denote the moment generating function of a random variable $X$. Now suppose that the following expression holds: for a given $a>0$ \begin{align} M_X(t) = 2 E \left[ e^{tX} \Phi( aX-t) \right], \forall t \in \mathbb{R} \end{align} where $\Phi$ is cdf of the standard normal.

Can we show that the only random variable that satisfies the above is Gaussian? The forward direction (i.e., evaluating with Gaussian) is not very difficult to show. However, the backward direction has been challenging.

**Proof of the direct part:** Choose $X$ to be zero mean Gaussian where variance $\sigma^2$ will be selected a bit later. Then, let $Z$ be standard normal, and note
\begin{align}
2 E \left[ e^{tX} \Phi( aX-t) \right]
&= 2 E \left[ e^{tX} 1_{\{Z \le aX-t\}} \right]\\
&= 2 E \left[ E[ e^{tX} 1_{\{Z \le aX-t\}}|Z] \right]\\
&= 2 E \left[ E \left[ e^{tX} |Z, \{Z \le aX-t\}\right] P[Z \le aX-t|Z] \right]\\
\end{align}

Now, the following is just the moment generating function of a truncated Gaussian \begin{align} E \left[ e^{tX} |Z, \{Z \le aX-t\}\right]= e^{\sigma^2\frac{t^2}{2}} \frac{1-\Phi(\frac{Z+t}{a\sigma}-\sigma t)}{1-\Phi(\frac{Z+t}{a\sigma})} \end{align} Also note that $P[Z \le aX-t|Z]=1-\Phi(\frac{Z+t}{a\sigma})$. Therefore, we arrive at \begin{align} 2 E \left[ e^{tX} \Phi( aX-t) \right]&=2 E \left[ e^{\sigma^2\frac{t^2}{2}} \left(1-\Phi(\frac{Z+t}{a\sigma}-\sigma t) \right)\right]\\ &=2 e^{\sigma^2\frac{t^2}{2}} E \left[ 1-\Phi \left(\frac{Z+t}{a\sigma}-\sigma t\right)\right]\\ &=2 e^{\frac{t^2}{2a}} E \left[1-\Phi ( \frac{Z}{\sqrt{a}}) \right] \text{ choose } \sigma=\frac{1}{\sqrt{a}}\\ &=2 e^{\frac{t^2}{2a}} \frac{1}{2} \text{ by symmetry}\\ &= e^{\frac{t^2}{2a}} \end{align}

Finally, note that the $X$ with $\sigma=\frac{1}{\sqrt{a}}$ has $M_X(t)= e^{\frac{t^2}{2a}}$

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