Double differentiation of characteristic function of Normal random variable

Question

Knowing that a one-dimensional random variable $\Gamma$ is Gaussian if it has the characteristic function $$\mathbb{E}\hspace{0.15cm}e^{i\xi\Gamma}=e^{im\xi-\frac{1}{2}\sigma^2\xi^2}\tag{1}$$ for some real numbers $m\in\mathbb{R}$ and $\sigma\geq0$. If we differentiate $(1)$ two times with respect to $\xi$ and set $\xi=0$, we see that $$m=\mathbb{E}\hspace{0.15cm}\Gamma\hspace{0.3cm}\sigma^2=\mathbb{V}\hspace{0.15cm}\Gamma\tag{2}$$

I cannot understand how to get to $(2)$ by double differentiating $(1)$ with respect to $\xi$, setting $\xi=0$.

If I differentiate r.h.s. of $(1)$ two times with respect to $\xi$ and I set $\xi=0$, I get $-m^2$.

Answer 1

This question, the other answer, and the comment discussion are muddled. For a random variable $X$ with finite second moment, with characteristic function $\varphi_X(t)=E[e^{itX}]$, the following are true: $$ \varphi_X(t)=1+ iE[X]t -\frac{E[X^2]}2 t^2+o(t^2)\\E[X] =\mu = -i\varphi_X'(0)\\E[X^2] = \mu^2+\sigma^2= -\varphi_X''0).$$ If you define $\psi=\log\varphi_X$, then $$\psi(t)=\mu it-\sigma^2\frac{t^2}2+o(t^2)\\ \mu = -i \psi'(0)\\ \sigma^2 = -\psi''(0),$$ which can be connected with the previous equations by application of the chain rule and product rule: $$\varphi_X(t)=e^{\psi(t)}\\ \varphi_X'(t) = e^{\psi(t)} \psi'(t)\\ \varphi_X''(t) = e^{\psi(t)}( \psi'(t))^2 + e^{\psi(t)}\psi''(t)$$$$ \varphi_X'(0) = e^{\psi(0)} \psi'(0)=\psi'(0)=-i\mu\\ \varphi_X''(0) = e^{\psi(0)}( \psi'(0))^2 + e^{\psi(0)}\psi''(0)=-\mu^2-\sigma^2,$$ and so on.

Answer 2

I will treat $\mathbb{E}$ as a constant. $$\mathbb{E} \cdot e^{iξ \Gamma}=e^{imξ−\frac{1}{2}σ^2ξ^2}$$ So then $$\mathbb{E} \cdot e^{iξ \Gamma}=\frac{e^{imξ}}{e^{\frac{1}{2}σ^2ξ^2}}$$ Differentiating once w.r.t $ξ$,

$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot e^{imξ} - e^{imξ} \cdot 2 \cdot\frac{1}{2}\sigma^2 \cdot ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{\bigg( e^{\frac{1}{2}σ^2ξ^2} \bigg)^2}$$

Simplifying, we obtain: $$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{e^{σ^2ξ^2}}$$ Furthermore, $$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \bigg( im \cdot e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \bigg) }{\bigg (e^{\frac{1}{2} \cdot σ^2ξ^2} \bigg)^2}$$ So $$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{im \cdot e^{imξ} - e^{imξ} \cdot \sigma^2 ξ}{e^{\frac{1}{2} \cdot σ^2ξ^2}}$$ Which means that $$\mathbb{E} i\Gamma \cdot e^{iξ \Gamma} \cdot e^{\frac{1}{2} \cdot σ^2ξ^2} =e^{imξ} (im - \sigma^2 ξ)$$ Multiplying both sides by $e^{-im ξ}$, $$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im ξ \bigg)} =im - \sigma^2 ξ$$ Differentiating again w.r.t $ξ$ we obtain: $$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im ξ \bigg)} \cdot \frac{d}{dξ}\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im ξ \bigg)= - \sigma^2$$ Thus, $$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im ξ \bigg)} \cdot\bigg( i \Gamma + 2\frac{1}{2} \cdot σ^2ξ - im\bigg)= - \sigma^2$$ Finally, we obtain that if we differentiate the first equation twice by $ξ$ we obtain that: $$\mathbb{E} i\Gamma \cdot e^{ iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im ξ} \cdot\bigg( i \Gamma + σ^2ξ - im\bigg)= - \sigma^2$$ Now setting $$ξ=0,$$ We obtain that: $$\mathbb{E} i\Gamma \cdot e^{ i(0) \Gamma + \frac{1}{2} \cdot σ^2(0)^2 - im (0)} \cdot\bigg( i \Gamma + σ^2(0) - im\bigg)= - \sigma^2$$ So then $$\mathbb{E} i\Gamma \cdot e^{0} \cdot\bigg( i \Gamma- im\bigg)= - \sigma^2$$ Thus, $$\mathbb{E} i\Gamma \cdot\bigg( i \Gamma - im\bigg)= - \sigma^2$$ Simplifying our expression, $$\mathbb{E} i^2\Gamma^2 - \mathbb{E}i^2 \Gamma m= - \sigma^2$$ $$-\mathbb{E}\Gamma^2 + \mathbb{E}\Gamma m= - \sigma^2$$ Then $$\mathbb{E}\Gamma m= \mathbb{E}\Gamma^2- \sigma^2$$ $$ m= \frac{\mathbb{E}\Gamma^2- \sigma^2}{\mathbb{E}\Gamma}$$ So $$ m= \Gamma - \frac{\sigma^2}{\mathbb{E}\Gamma}$$ You should note that I treat $\mathbb{E}$ as a constant. If it is a function of $ξ$, I will edit my answer. But since RHS isn't dependent on $\mathbb{E}$, I can safely say that the second derivative of the RHS is \begin{align*} \frac{d^2}{dξ^2} \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) &= \frac{d}{dξ}\bigg[ \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \cdot \bigg( im - σ^2ξ \bigg) \bigg] \\ &= \frac{d}{dξ}\bigg( im \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - σ^2ξ \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \\ &= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \bigg[ \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} + \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg] \\ &= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \\ &= e^{imξ−\frac{1}{2}σ^2ξ^2} \cdot \bigg[ im \big( im - \sigma^2 ξ \big) - \sigma^2ξ \big( im - \sigma^2 ξ \big) - \sigma^2 \bigg] \end{align*} Now setting $ ξ=0$ we obtain that: $$\frac{d^2}{dξ^2} \big( e^{imξ−\frac{1}{2}σ^2ξ^2} \big) \bigg|_{ξ=0} = i^2m^2 - \sigma^2 = -m^2 - \sigma^2$$