Distribution of Fractional Polynomial of Random Variables

liuch37

New member
Hi all,

I would like to find the distribution (CDF or PDF) of a random variable Y, which is written as

Y=X_1*X_2*...X_N/(X_1+X_2+...X_N)^N.

X_1, X_2,...X_N are N i.i.d. random variables and we know they have the same PDF f_X(x).

I know this can be solved by change of variables technique and use multi-dimensional integration. But is there a easier way to deal with it?

Thanks.

chisigma

Well-known member
Hi all,

I would like to find the distribution (CDF or PDF) of a random variable Y, which is written as

Y=X_1*X_2*...X_N/(X_1+X_2+...X_N)^N.

X_1, X_2,...X_N are N i.i.d. random variables and we know they have the same PDF f_X(x).

I know this can be solved by change of variables technique and use multi-dimensional integration. But is there a easier way to deal with it?

Thanks.
There is a 'milestone' that allows You to solve the problem : if $X_{1},X_{2},...,X_{n}$ are independent r.v. with p.d.f. $f_{1}(x),f_{2}(x),...,f_{n}(x)$, then the r.v. $X= X_{1} + X_{2} + ...+ X_{n}$ has p.d.f....

$$f(x) = f_{1} (x) * f_{2} (x) * ... *f_{n} (x)\ (1)$$

... where '*' means convolution. Now let's introduce for any r.v. $X_{i}$ the auxiliary r.v. $\Lambda_{i} = \ln |X_{i}|$ with p.d.f. $\lambda_{i} (x)$. Because is...

$$\ln |Y| = \ln |X_{1}| + \ln |X_{2}| + ... + \ln |X_{n}| - n\ \ln |X_{1} + X_{2} + ... + X_{n}|\ (2)$$

... You can apply (1). The only criticity is the fact that the r.v. $\sum_{i} \ln |X_{i}|$ and $n \ln |\sum_{i} X_{i}|$ are not independent.

An example: if the $X_{i}$ are all uniformely distributed in [0,1], what is the p.d.f. of $Y = \sum_{i} \ln X_{i}$?... If X is uniformely distributed in [0,1], then...

$$P \{ \ln X < - x \} = \int_{e^{- x}}^{1} d\ \xi = 1 - e^{- x}\ (3)$$

Deriving (3) we find that the r.v. $\ln X$ has p.d.f. $e^{-x}$, which has Laplace Transform...

$$\mathcal{L} \{e^{-x} \} = \frac{1}{1 + s}\ (4)$$

Now we can find that the p.d.f. of Y ...

$$\lambda (x) = \mathcal {L}^{-1} \{ \frac{1}{(1+s)^{n}}\} = \frac{x^{n-1}}{(n-1)!}\ e^{-x}\ (5)$$

Kind regards

$\chi$ $\sigma$