I am trying to find the PDF of product of two random variables with the following distributions,

       f1 = (t/(T Pi r Sqrt[1-((x1^2)/(r^2))])) + (1-t/T) DiracDelta[x1]; 
domain[f1] = {x1, -r, r} && {T> t > 0, r > 0};

f1 is mixed distribution where, x1 follows a continuous distribution with probability t/T and rest of the time(1-t/T) the variable, x1=0.

       f2  = (2 F^(2/n)/((a^2) n)) x2^(-(2 + n)/n);   
domain[f2] = {x2, F*(a^-n), Infinity} &&  {a > 0, F > 0, n > 0};

I used the TransformProduct function in MathStatica to generate the PDF of the product x1*x2

f3 = TransformProduct[{f1, f2}, y]

I get the following result pdf of f3

However, when I integrate f3, I get a numerical result equal to t/T, not 1 whereas both f1 and f2 integrates to 1. I feel like there is missing Dirac-delta in the f3 with area (1-t/T) like the one I have in f1. Am I doing anything wrong? Thanks.

  • 1
    $\begingroup$ I don't have a TransformProduct function in Mathematica 10.4. Are you using (the excellent product) mathStatica ? If so, you should mention that. $\endgroup$
    – JimB
    Apr 6, 2016 at 20:36
  • $\begingroup$ Thanks @JimBaldwin I have mentioned now. $\endgroup$
    – Hossain
    Apr 6, 2016 at 21:49
  • $\begingroup$ Upgrade you math: you use the DiracDelta distribution as an atomic measure. See en.wikipedia.org/wiki/Dirac_delta_function . Unfortunately, that topic in the article en.wikipedia.org/wiki/Probability_density_function is not correctly written . $\endgroup$
    – user64494
    Mar 27, 2019 at 19:15

1 Answer 1


This is an unusual and interesting question. This is a bit obscured by the many parameters that make it difficult to see the wood for the trees.

Short Answer

Your first pdf, which has the DiracDelta function, is a mixed discrete / continuous random variable. To make this clear, if say $r = 1$, $t = 1$ and $T = 2$, then your mixed pmf/pdf f1 is:

    f1 = 1/(2Pi Sqrt[1-x1^2]) + DiracDelta[x1]/2

This is just a weighted average mix of an ArcSine random variable and 0. It could alternatively (and perhaps more naturally) be written as:

    f1 = If[ x1 == 0, 1/2, 1/(2Pi Sqrt[1-x1^2])]

which would be plotted as:

(source: tri.org.au)

By contrast, the TransformProduct function states in the Help system:

   `Density f and density g can be continuous or piecewise continuous.`

This defn does not include mixed cont/discrete random variables.

Using a DiracDelta function is an interesting way of trying to express a discrete model in a continuous space, but I am not sure that it is appropriate. One cannot even seem to plot it appropriately, though Integrate works.

Even though your mixed cont/discrete model is 'incompatible' with the restriction to continuous variables, all is not lost! One can easily provide a workaround, simply by formulating the problem slightly differently ...

Full Answer

Let us deconstruct your $X_1$ random variable back to its separate continuous and discrete parts. Then:

$$ X_1 = \begin{cases}0 & \text{with probability } p \\ Z & \text{with probability } 1-p \end{cases}$$

where $p = \frac{t}{T}$, and random variable $Z$ has pdf $f(z)$:

(source: tri.org.au)

Next, $X_2$ has pdf, say $f_2(x_2)$:

(source: tri.org.au)

The desired product of random variables is:

$$ Y = X_1 X_2 = \begin{cases}0 & \text{with probability } p \\ Z*X_2 & \text{with probability } 1-p \end{cases}$$

Then, the mixed pdf of $Y$ is:

$$ \text{pdf}(Y) = \begin{cases}p & \text{if } y = 0 \\ (1-p) h(z * x_2) & \text{if } y \neq 0 \end{cases}$$

where $h(z * x_2)$ denotes the pdf of the product $Z X_2$. The latter ($Z$ and $X_2$) are both continuous random variables, so you can use the TransformProduct function from the mathStatica package to find that $h(z * x_2)$ is:

(source: tri.org.au)

... and we are all done.

Here is a plot of $h(.)$ (the continuous part) when $F=1$, $n = 1$, $a = 2$, $r=1$:

(source: tri.org.au)

All that happens when you embed $h(.)$ into the mixed model $\text{pdf}(Y)$ above is that the continuous density gets scaled down by $(1-p)$, and there will be a discrete mass $p$ at 0.

  • $\begingroup$ Thank you very much @wolfies. I appreciate the help! $\endgroup$
    – Hossain
    Apr 7, 2016 at 15:45
  • $\begingroup$ See my comment to the question. $\endgroup$
    – user64494
    Mar 27, 2019 at 19:16

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