# Convolution of two distribution functions

I have two functions;

f[x_] = (1/k) Exp[-x/k] ;
g[x_] = (1/p) Exp[-x/p] ;


How I can convolve them?

In Mathematica for convolving two functions we have this function:

Convolve[f, g, x, ??]


But I don't know how to do the convolution with the above function for my case.

before I ask this question I searched in detail but I didn't find what I need.

Note: the solution of the above functions is:

(1/k) Exp[-x/k]  *  (1/p) Exp[-x/p]  =  (1/(k-p)) ( Exp[-x/k]  - Exp[-x/p] )


where * is the convolution.

So, how to do this convolution in Mathematica?

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Convolve[f, g, x, y] gives zero (y is the convolution dummy variable) –  Nasser Sep 10 '13 at 18:08
@Nasser You don't happen to have used the OP definitions of the functions? They are not written with the correct Mathematica syntax. Should have been f = (1/k) Exp [-x/k];g = (1/p) Exp [-x/p]; –  Sjoerd C. de Vries Sep 10 '13 at 18:38
You need to use proper Mathematica syntax; the exponential function is Exp[]. Next issue is that the integral will not converge on [-Infinity, Infinity]. –  b.gatessucks Sep 10 '13 at 18:38
@SjoerdC.deVries I just copied what OP wrote and did not even notice the exp vs. Exp ;) , oh boy, now I look at it, I see more syntax errors , using "(" vs. "[" . I need to go make some strong coffee –  Nasser Sep 10 '13 at 18:45

The functions do not have a finite area, so they cannot be real distributions as your title claims they are.

Let's change them a bit so they have area 1.

f[x_] = (1/k) Exp[-x/k] UnitStep[x];
g[x_] = (1/p) Exp[-x/p] UnitStep[x];

Integrate[f[x], {x, -∞, ∞}]


ConditionalExpression[1, Re[1/k] > 0]

The convolution:

Convolve[f[x], g[x], x, y]


which equals (well apart from the unit step) what you were expecting.

Since your title mentions convolution of distributions let's explore that route as well. A convolution of two probability distributions is defined as the distribution of the sum of two stochastic variables distributed according to those distributions:

PDF[
TransformedDistribution[
x + y,
{
x \[Distributed] ProbabilityDistribution[f[x], {x, -∞, ∞}],
y \[Distributed] ProbabilityDistribution[g[x], {x, -∞, ∞}]
}
],x
]


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