Suppose that $X \sim N (0,1)$ and $Y \sim \operatorname{Exp}(1)$ are independent random variables. Prove that $X\sqrt {2Y}$ has a standard Laplace distribution.

My first approach was the following:

$F_Z(z)=P (Z\leq z) = P (X\sqrt {2Y}\leq z) = P (X\leq \frac {z}{\sqrt {2Y}} ) =$ $$=\int_{\Re} P (X\leq \frac {z}{\sqrt {2Y}}\mid Y=y)\cdot f_{Y}(y) \, dy = \int_{\Re} F_{X}(\frac {z}{\sqrt {2y}})\cdot f_{Y}(y) \, dy$$

Am I on the right track? Have I made any mistakes on the way? If not, how can I proceed from there to determine $f_Z$?

My second approach was the following (change of variable method):

Set $Z=X\sqrt {2Y}$ and $W=Y$, then, $f_{Z,W}=\frac {f_{X,Y}(\frac z {\sqrt {2w}},w)}{\sqrt {2w}}=\frac {f_X(\frac z {\sqrt {2w}})\cdot f_Y(w)}{\sqrt {2w}}$, and:

$$f_Z(z)=\int_{\Re} f_{Z,W}(z,w)\,dw=\int_{\Re}\frac {f_{X}(\frac {z}{\sqrt {2w}})\cdot f_{Y}(w)}{\sqrt {2w}} \, dw$$

Have i made any mistakes? And again, if not, how can I proceed? Cant find any way to solve the integral above, making me think that i might have made a mistake on the way. Thanks !


2 Answers 2


Incomplete answer, but too long for a comment:

If one can show

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{x^4+z^2}{2x^2}} \mathop{dx} = e^{-z},$$

then we can get the answer by noting that for $z \ge 0$, \begin{align} P(|X\sqrt{2Y}| \le z) &= \int_{-\infty}^\infty F_Y(\frac{z^2}{2x^2}) \cdot f_X(x) \mathop{dx} \\ &= \int_{-\infty}^\infty (1-e^{-\frac{z^2}{2x^2}}) \cdot \frac{e^{-x^2/2}}{\sqrt{2\pi}} \mathop{dx} \\ &= 1 - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{x^4+z^2}{2x^2}} \mathop{dx} \\ &= 1 - e^{-z}. \end{align} Then, using the fact that $X\sqrt{2Y}$ is a symmetric random variable, you can recover the CDF of the Laplace distribution.

However, I do not know how to verify the above integral...

Edit: Flowsnake has computed the integral, thanks! I am still curious if there is an easier approach to the original question.

  • $\begingroup$ Per Maple, your first integral is as stated. $\endgroup$ Jul 30, 2017 at 1:11

To compute the integral given by @angryavian, first note that $-\frac{1}{2}\left(x^2 + \frac{z^2}{x^2}\right) = -\frac{1}{2}\left(x - \frac{z}{x}\right)^2 - z$. Then,

$$ \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty e^{-\frac{1}{2}\left(x^2 + \frac{z^2}{x^2}\right)}dx = \frac{e^{-z}}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty e^{-\frac{1}{2}\left(x - \frac{z}{x}\right)^2}dx = \frac{e^{-z}}{\sqrt{2\pi}}2I \;\;\;\; (**) $$

where $I = \displaystyle\int\limits_{0}^\infty e^{-\frac{1}{2}\left(x - \frac{z}{x}\right)^2}dx$. Making the substitution, $u = \dfrac{z}{x}$ gives, $I = \displaystyle\int\limits_{0}^{\infty}\frac{z}{u^2}e^{-\frac{1}{2}\left(u - \frac{z}{u}\right)^2}du$. Then, adding the integrals together gives,

$$ 2I = \displaystyle\int\limits_{0}^{\infty}\left(1+\frac{z}{u^2}\right)e^{-\frac{1}{2}\left(u - \frac{z}{u}\right)^2}du = \displaystyle\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}v^2}dv = \sqrt{2\pi}$$

where the second to last equality follows from the substitution, $v =u - \dfrac{z}{u}$. Plugging this into $(**)$ gives the result.


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