# Function of a random variable

Here is a probability density function (PDF)

pdfq[q_] = C q^(4 n ν - 1) (1 - q)^(4 n μ - 1)


$$P_q(Q=q) = C q^{4 n\nu - 1} (1 - q)^{4 n \mu - 1}$$

I'd like to find the PDF of $l$ (noted $P_l(L=l)$), which is a function of $q$.

f[q_] = 2 q s


$$f(q)=2\cdot q\cdot s$$

What is the PDF of l?

Here is my try

I think $f(q)$ is an invertible, differentiable and continuously increasing function of $q$ and therefore:

$$P_l(L=l) = P_q(f^{-1}(l))\left|\frac{df^{-1}(l)}{dl}\right|$$

according to wiki, where $f^{-1}(..)$ is the inverse function of $f(..)$.

Calculating this with Mathematica gives:

In[1]:= pdfq[q_] =
C E^(4 n s q ) q^(4 n ν - 1) (1 - q)^(4 n μ - 1)

Out[1]= C E^(4 n q s) (1 - q)^(-1 + 4 n μ) q^(-1 + 4 n ν)

In[2]:= f[q_] = 2 q s

Out[2]= 2 q s

In[3]:= Solve[l == f[q], q]

Out[3]= {{q -> l/(2 s)}}

In[4]:= finverse[l_] = l/(2 s)

Out[4]= l/(2 s)

In[5]:= Absdfinversedl[l] = Abs[Integrate[finverse[l], l]]

Out[5]= 1/4 Abs[l^2/s]

In[7]:= pdfl[l_] = pdfq[finverse[l]] Absdfinversedl[l]

Out[7]= 2^(-1 - 4 n ν) C E^(
2 l n) (1 - l/(2 s))^(-1 + 4 n μ) (l/s)^(-1 + 4 n ν)
Abs[l^2/s]

In[8]:= pdfl[0.5] /. C -> 1 /. n -> 1000 /. μ -> 0.02 /. ν ->
0.05 /. s -> 0.1 // N

Out[8]= -1.554651720194871*10^527


But the result is obviously not correct as I get negative probability densities. What am I doing wrong? Does the problem have to do with my constant of integration $C$? Thanks a lot for your help.

First look into the documentation to see if your distribution is already defined. If it is not, you can define your own distribution based on its PDF or CDF using ProbabilityDistribution.

dist=ProbabilityDistribution[ C q^(4 n Nu - 1) (1 - q)^(4 n Mu - 1), {q, 0, 1}, ...]


(This is incomplete, you need to define the ranges of the parameters)

Whatever ranges you variables and parameters are, you could test it by obtaining Mean[dist] or PDF[dist,x] and checking that the there are no errors.

Now you have your original distribution, you want to transform it using TransformedDistribution .

tDist = TransformedDistribution[2 q s, q \[Distributed] dist[Nu, Mu]]


Now the PDF of your new transformed distributions would be PDF[tDist,x]

• Thank you. I have a hard time to understand how to deal with pdf in mathematica. I ask another question here. – Remi.b Oct 21 '14 at 20:53
• Does it also makes sense if the expression is a function of the parameters of the distributions? Something like dist = TransformedDistribution[2 µ a, a \[Distributed] NormalDistribution[µ, \[Sigma]]] or for my example, something like tDist = TransformedDistribution[2 q Nu, q \[Distributed] dist[Nu, Mu]], (where I replaced $s$ by $Nu$)? Thanks a lot! – Remi.b Oct 28 '14 at 20:14