Let me answer this, because I think you are asking the wrong question.
Assume that you have two random variables, one normal (call it W, for wheel) and one exponential (call it f, for friction.)
You want to compute the product of those random variables, then to know what the distribution of the product would be. Much of the time, that task is impossible to anser exactly. Thus, given some genral nonlinear function of multiple random variables, what is the distribution of some general function f(X,Y,...).
Sometimes, it is trivially easy. For example,the sum of any number of normal (Gaussian) ditributions is also Normally distributed. it is even easy to compute the mean and variance of the resulting Normal sum. Great. But the problem is, we get spoiled. And in fact, most nonlinear functional transformations of distributions are not so easily dealt with. The general field that deals with these questions is sometimes called Statistical Tolerancing, also Tolerance Analysis.
Mathematically, one can indeed write the desired distribution as sort of a convolution integral, involving the the corresponding PDFs. But the problem is, it won't usually be anything useful. There simply are not that many distributions out there that will represent the distribution of any general function of any set of random variables.
In the case of the desired product distribution, you can find it here
On that page, you will find the distribution shown as an integral.
Pz(x) = int(Px(x) * Py(z/x) *1/abs(x),[-inf,inf])
Since one of the distributions is an exponential, assume that is x. So we need only integrate from 0 to inf. We would then have
Px(x) = lambda*exp(-lambda*x)
And
Py(y) = 1/(sigma*sqrt(2*pi))*exp(-(y - mu)^2/(2*sigma^2))
But sadly, it does not appear the result is anything very usful. I'm not even positive the integral will have an analytical solution. But, just for kicks, lets see what the symbolic toolbox can do with the product of a standard exponential and a standard normal from that expression. So we would have
lambda = 1
sigma = 1
mu = 0
The integral would be written as:
syms x z
int(exp(-x)*1/sqrt(2*pi)*exp(-(z/x)^2/2)/x,x,[0,inf])
ans =
piecewise(angle(1/z^2) in Dom::Interval([-pi/2], [pi/2]), (1125899906842624*meijerG([1/2, 1, 1], [], [], [], 8/z^2))/(5644425081792261*pi^(1/2)), ~angle(1/z^2) in Dom::Interval([-pi/2], [pi/2]), int((2251799813685248*exp(-x)*exp(-z^2/(2*x^2)))/(5644425081792261*x), x, 0, Inf))
So, way less interesting that we want.
In practice, one uses other methods to approximate the desired distribution. First though, can we recognize the distribution? For example, is the product also normally distributed?
x = exprnd(1,1,1e8);
y = randn(1,1e8);
mean(x.*y)
ans =
-0.00017924
var(x.*y)
ans =
2.0007
skewness(x.*y)
ans =
-0.0017322
kurtosis(x.*y)
ans =
18.055
Hmm. I'd expect the odd moments to approch zero for large sample sizes. Ans 1e8 is quite large as sample sizes go.
It looks like the variance of the product seem to approach 2. The problem is the 4th moment. A normal has a Kurtosis of 3. That means the product distribution is not normally distributed, with differently shaped tails compared to a Normal distribution. We can see that from a histogram.
hist(x.*y,1000)
What distribution is this?
[r,type] = pearsrnd(0,sqrt(2),0,18,0,0)
r =
[]
type =
7
Which pearsrnd tells us seems to approximately a Student-t, as close as it can come. Not exactly, but a Student's t would be close.
Type 7: Student's t location-scale
Now, lets try it. I'll use a smaller sample size to not get the distribution fitter too upset.
x = exprnd(1,1,1e6);
y = randn(1,1e6);
z = (x.*y)';
D = fitdist(z,'tlocationscale')
D =
tLocationScaleDistribution
t Location-Scale distribution
mu = 0.000146401 [-0.000891219, 0.00118402]
sigma = 0.407176 [0.40559, 0.408768]
nu = 1.22331 [1.21773, 1.22891]
histogram(z',100,'normalization','pdf')
hold on
fplot(@D.pdf)
The t actually looks alright. Not perfect. But decent.
As we saw above, IF this were a Student's t, then we could back out a degrees of freedom. I.e., if the variance is 2 then the corresponding degrees of freedom would be nu/(nu-2). Solving for nu, we get 4 effective degrees of freedom. But a t with 4 degrees of freedom has a Kurtosis of inf.
Oh well, the point of all this is you won't get an exact pdf out of this product of an exponential and a Gaussian, but it looks as if you will do alright if you treat it as a Student's t.
