# Product of a normal distribution and an exponential distribution?

I am trying to find the product of normal distribution and Exponential distribution (Both are independent). Could we do analytically?

• Welcome to MMA StackExchange. This question doesn't seem to be related to the software Mathematica? Have you tried anything yet ? – Dunlop Jul 29 at 9:12
• Wolfram is unable to find this one$f(z)=\int_0^{\infty}\frac{\lambda}{x\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}-\frac{\lambda z}{x}}$. – dtc348 Jul 29 at 9:14
• Mathematica finds PDF[TransformedDistribution[ x*y, {x [Distributed] NormalDistribution[0, [Sigma]], y [Distributed] ExponentialDistribution[[Lambda]]}], t] in terrms of MeijerG function $$\frac{\lambda G_{0,3}^{3,0}\left(\frac{t^2 \lambda ^2}{8 \sigma ^2}| \begin{array}{c} 0,0,\frac{1}{2} \\ \end{array} \right)}{2 \sqrt{2} \pi \sigma } .$$ There are wishes and there is reality. – user64494 Jul 29 at 9:50
• @user64494 I am sorry, I don't know, how to change above integration function in to Meijer G function. Do we analytically change into Meijer G function? – dtc348 Jul 29 at 9:55
• Possible duplicate of the linked question. – lirtosiast Jul 29 at 13:25

Clear["Global*"]

distx = NormalDistribution[0, σ];

disty = ExponentialDistribution[λ];

assume = DistributionParameterAssumptions[distx] &&
DistributionParameterAssumptions[disty]

(* σ > 0 && λ > 0 *)

(cdf[z_] =
Probability[
x*y <= z, {x \[Distributed] distx,
y \[Distributed] disty}]) //


(pdf[z_] = Assuming[assume, D[cdf[z], z]]) //


The expressions for z < 0 and z > 0 are identical

Assuming[assume,
Equal @@ (Simplify[pdf[z], #] & /@ {z < 0, z > 0}) //
Simplify]

(* True *)


Consequently, simplify the expression to

(pdf[z_] = Simplify[pdf[z], z > 0]) //


Alternatively,

distz = TransformedDistribution[x*y,
{x \[Distributed] NormalDistribution[0, σ],
y \[Distributed] ExponentialDistribution[λ]}];

(pdf2[z_] = PDF[distz, z]) // TraditionalForm


Although it is difficult to show that the expressions for the PDF are equivalent, the numeric difference between the two expressions is zero:

Plot[pdf[z] - pdf2[z] /. {σ -> λ}, {z, -4, 4},
PlotRange -> {-10^-6, 10^-6},
WorkingPrecision -> \$MachinePrecision]
`

• Nice to see that MMA pulls this off. Springer showed that such a product is a MeijerG function back in the 70s... – ciao Jul 29 at 21:37