# product of two probability density functions

I have on one hand a z variable following a standard normal distribution N(0,1) and on the other hand a variable X following a Nakagami distribution. I am supposed to find a Gaussian distribution when I take the product of the two distributions described above. I can't find it as I am left with an expression resembling a gamma distribution inside the integral. Has anyone come across that kind of question? Are there different kinds of Nakagami distributions?

• Are the parameters of the Nakagami distribution fixed and given, or do they have to be chosen arbitrarily? Apr 29 '20 at 21:56
• I mentioned Nakagami as I have the square root of a gamma random variable. I am not sure how to select the Nakagami parameters Apr 29 '20 at 22:22
• Chosen arbitrarily I guess Apr 29 '20 at 22:39
• Your title says you're multiplying distributions, which I would normally take to mean "multiplying distribution functions" but could perhaps mean "multiplying pdfs". However, reading your body text I wonder if you might actually mean to ask about the distribution when multiplying the random variables. Can you make it explicit what is actually being multiplied please? Apr 30 '20 at 12:47
• Yes I actually mean multiplying pdfs Apr 30 '20 at 13:15

If $$Y=b_0+Z/\sqrt{\lambda \tau}$$ where $$Z\sim N(0,1)$$ and $$\lambda$$ has either a gamma distribution (with parameters $$a$$ and $$b$$) or a Nakagami distribution (with parameters $$m$$, and $$w$$), then Mathematica (and not my limited algebraic skills) finds the following pdf's for $$\tau>0$$ and assuming that $$Z$$ and $$\lambda$$ are independent:

d1 = TransformedDistribution[b0 + z/Sqrt[n \[Tau]], {z \[Distributed] NormalDistribution[0, 1],
pdf1 = PDF[d1, z]


$$\frac{2^a \sqrt{b} \sqrt{\tau } \Gamma \left(a+\frac{1}{2}\right) \left(b \tau (b_0-z)^2+2\right)^{-a-\frac{1}{2}}}{\sqrt{\pi } \Gamma (a)}$$

d2 = TransformedDistribution[b0 + z/Sqrt[n \[Tau]], {z \[Distributed] NormalDistribution[0, 1],
n \[Distributed] NakagamiDistribution[m, w]}];
pdf2 = PDF[d2, z]


$$\frac{\sqrt{\tau } \sqrt[4]{\frac{w}{m}} \left(2 \Gamma \left(m+\frac{1}{4}\right) \, _1F_1\left(m+\frac{1}{4};\frac{1}{2};\frac{w (b_0-z)^4 \tau ^2}{16 m}\right)-\tau (b_0-z)^2 \sqrt{\frac{w}{m}} \Gamma \left(m+\frac{3}{4}\right) \, _1F_1\left(m+\frac{3}{4};\frac{3}{2};\frac{w (b_0-z)^4 \tau ^2}{16 m}\right)\right)}{2 \sqrt{2 \pi } \Gamma (m)}$$

Maybe you are directly integrating the product of the pdf's of $$Z$$ and $$\lambda$$ to obtain the pdf of $$Y$$ but it's all about how you do the integration which is why I think your title is misleading. I have chosen the lazy way (and most efficient for me).

As an example consider $$b_0=0$$, $$a=1$$, $$b=2$$, $$m=1$$, $$w=2$$, and $$\tau=1$$:

Plot[{pdf1 /. {b0 -> 0, m -> 1, w -> 2, \[Tau] -> 1},
pdf2 /. {b0 -> 0, a -> 1, b -> 2, \[Tau] -> 1}},
{z, -5, 5}, WorkingPrecision -> 30,
PlotLegends -> {"\[Tau]=1 and Nakagami[1,2] distribution",
"\[Tau]=1 and Gamma[1,2] distribution"}]


• yes, I am directly integrating the product of the pdf's of 𝑍 and 𝜆 to obtain the pdf of 𝑌.Since I can't upvote posts yet, I allow myself to comment a "thank you". Apr 30 '20 at 18:34
• Thanks for the accept. If you could, adding in the details into you question would be great. My answers were more along the lines of guesses as to what I thought you wanted and this forum is for more of a long term repository of questions and answers. So it would help future inquires when folks search this site.
– JimB
Apr 30 '20 at 19:41