MHB Distribution of Fractional Polynomial of Random Variables

AI Thread Summary
The discussion focuses on finding the distribution of a random variable Y defined as Y = X_1*X_2*...*X_N/(X_1+X_2+...+X_N)^N, where X_1, X_2,...,X_N are independent and identically distributed random variables. The initial approach suggested involves using a change of variables and multi-dimensional integration, but participants are exploring simpler methods. A key insight shared is the use of auxiliary random variables, specifically the logarithm of the X variables, to facilitate the analysis. An example is provided with uniformly distributed X variables, leading to the derivation of the probability density function for Y. The conversation highlights the complexity of dependency among transformed variables and the potential for convolution techniques in solving the problem.
liuch37
Messages
1
Reaction score
0
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.
 
Physics news on Phys.org
liuch37 said:
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

$\chi$ $\sigma$
 
Back
Top