Normal distribution can not be transformed into Laplace via additive transformation

Question

I am trying to find different methods of showing that there exists no random variable $V$ that can additively transform standard normal random variable into Laplace random variable.

Formally, we would like to show that there is no random variable $V$ independent of $N$(standard normal) such that \begin{align} W=V+N, \end{align} where $W$ is zero mean and with Laplace parameter $b\ge 1$. We want to show this is impossible for all $b\ge 1$.

Proof 1 (Via Characteristic Functions): I have a proof via characteristic functions that goes as follows \begin{align} \phi_W(t)=\phi_V(t) \phi_N(t) \rightarrow \phi_V(t)=\frac{\phi_N(t)}{\phi_W(t)} \end{align} now we know that $\phi_N(t)=e^{-\frac{t^2}{2}}$ and $\phi_W(t)=\frac{1}{1+b^2 t^2}$, so \begin{align} \phi_V(t)= \frac{e^{\frac{t^2}{2}}}{1+b^2 t^2} \end{align} Clear this can not be a characteristic function since there are values of $t$ for which $\phi_V(t) >1$.

Proof 2 (Via analyticity after convolution) This proof was suggested by D.Thomine (see comments below). Since convolution 'improves' regularity the pdf of $W$ must be analytic, however, the pdf of Lapalace distribution is not analytic at zero.

My question: What would be some other methods of showing this?

I was also thinking of the following approach. We know that \begin{align} c_1 e^{-|w|/b} &= E[ c_2 e^{-|X+w|^2/2} ] \end{align}

Can we show that \begin{align} E[ c_2 e^{-|X+w|^2/2} ] \le c_3 e^{-w^2/2} \end{align} this would lead to a contraditiction since $e^{-w^2/2}$ decays faster than $e^{-|w|/b}$.

This question is related to something I asked here .

Looking forward to your solutions. Thank you.

Answer 1

What you use to solve this problem is that the Fourier transform of a standard Gaussian decays much faster than the Fourier transform of $e^{-|x|}$. If we handwave a little, this means that the Gaussian is much smoother than $e^{-|x|}$, which is something we can exploit directly.

The main idea is classical : adding an independent real-valued random variable is equivalent, at the distribution level, with convolution the densities, and convolution improves smoothness. This is often used when we want to deal with smooth distribution; we just need to add some random noise to smooth any initial distribution.

Proposition

Let $N$ be a standard normal random variable. Let $V$ be a real-valued random variable, independent from $N$. Then the density of the distribution of $N+V$ has an entire version.

Proof

Let $\mu$ be the distribution of $V$. Then $(\mathbb{R}, \mathcal{B}, \mu)$ is a measure space.

For $M >0$, let $G_M := \{z \in \mathbb{C}: \ |Im(z)| <M\}$. Then $G$ is open, and for $z =: x+iy \in G$,

$$\left| \frac{1}{\sqrt{2 \pi}}e^{-\frac{z^2}{2}} \right| = \frac{1}{\sqrt{2 \pi}}e^{\frac{y^2-x^2}{2}} \leq \frac{1}{\sqrt{2 \pi}}e^{\frac{M^2}{2}}.$$

For $z \in G_M$ and $\omega \in \mathbb{R}$, let $f(z,\omega) := \frac{1}{\sqrt{2 \pi}}e^{-\frac{(z-\omega)^2}{2}}$. Then $f$ is bounded, holomorphic in the first variable, and measurable in the second. Hence, using complex differentiation under the integral, the function :

$$F: z \mapsto \int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi}}e^{-\frac{(z-\omega)^2}{2}} d \mu (\omega)$$

is holomorphic on $G_M$. Since this is true for all $M>0$, the function $F$ is holomorphic on $\mathbb{C}$, and thus, entire. But $F$ is exactly the distribution of $N+V$.

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All we need to conclude is the observation that the density of a Laplace random variable is never smooth at $0$.