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I am trying to find the joint pdf of the random variable z which is the summation of $n$ i.i.d random variables i.e.:

$$Z= X_1 + X_2+ ....+ X_n$$

Where all the $X_i$'s are iid random variables with $Px= C/x^2$, $a \le X \le b$
$C=1/624 $, $a=1/625 $, $b=1$.

I tried the Jacobian method for $n=3$ but I ended up with integration of the form

$Pz= \int_{a2}^{b2} \int_{a1}^{b1} 1/(z_1)^2 * 1/(z_2 - z_1)^2 *1/(z - z_2)^2 dz_1 dz_2\;$, for $a\le z \le 3b$.

Next I tried the characteristic function approach. The idea is: Since $X_i$'s are iid then the c/c function of $Z$ is just the c/c function of $X$ raised to the power $n$. once the c/c function of $Z$ is found I can apply Fourier transform and get the pdf of z.

When I tried to get the c/c function of but immediately I got this integration:

$\phi_x= \int_{a}^{b} C*e^{i*t*x}/x^2 dx$

Any idea how to solve it?

I used wolfram alpha but it gave me the answer in terms of Ei(). I saw couple of nice tricks regarding contour integration and complex analysis that were used during finding the c/c function of a normal random variable and I thought some one might have an idea about how to apply such tricks to this pdf as well. I tried to find the c/c function of the normal R.V using Mathematica but it didn't give me the typical $\phi_{normal}=e^{-t^2/2}$ that is why I have suspensions about the Ei() which he gave me as a solution.

Thank you very much. All comments are welcome (the whole point of the post is to discuss and get more perspectives).

Regards.

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I don't know if this will help you but your density represents a truncated from above Pareto distribution.

PROOF: The density and the cumulative distribution function of the Pareto distribution are

$$f_X(x) = \frac {\gamma x_{\min}^{\gamma}}{x^{\gamma +1}}\qquad F_X(x) = 1- \left(\frac {x_{\min}}{x}\right)^{\gamma} \qquad \gamma >0,\;\; x_{\min}>0,\;\;x\in [x_{\min}, \infty]$$

To match it to your case, we must have $\gamma =1$ and $x_{\min} = a$.

The support is truncated from above at $b$ so the truncated density, denote it $\hat f$ is

$$\hat f_X(x) = \frac {f_X(x)}{F_X(b)} $$ which in your case gives (since also $b=1$) $$\hat f_X(x) = \frac {a/x^2}{1- a} = \frac {1/625}{1-1/625}\frac {1}{x^2} = \frac {1}{624}\frac {1}{x^2} = C/x^2$$ ... which is what I wanted to show. Although for $\gamma =1$ the moments of the untruncated Pareto distribution do not exist, this is not so for a truncated version. For example

$$E(X) = \int_a^1x\frac {a/x^2}{1- a}dx = \frac {a}{1- a}\int_a^1\frac 1xdx = \frac {a}{1- a} (\ln1 - \ln a) = \frac {-a\ln a}{1- a} $$.

Maybe this characterization can lead you somewhere.

ADDENDUM: *The moment generating function of the truncated Pareto Distribution*

In general, we have, as an instance of a Mellin transform, $$ \int_a^{\infty}x^{s-1}e^{-\xi x}dx = \xi^{-s}\Gamma (s, \xi a)\qquad \xi>0,\; a>0$$ where $\Gamma (\;, \,)$ is the upper incomplete gamma function. In the present case, $s=-1$ so setting $t\equiv -\xi$ we can express the moment generating function of the specific truncated Pareto distribution as

$$MGF_X(t) = \int_a^1e^{tx}\frac {a}{1- a}x^{-2}dx = \int_a^{\infty}e^{tx}\frac {a}{1- a}x^{-2}dx - \int_1^{\infty}e^{tx}\frac {a}{1- a}x^{-2}dx$$ $$= -t\Gamma (-1, -t a)\frac {a}{1- a} + t\Gamma (-1, -t )\frac {a}{1- a}$$ $$=\frac {at}{1- a}\Big(\Gamma (-1, -t ) - \Gamma (-1, -t a) \Big)$$

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  • $\begingroup$ Thank you very much, I will try to search in that direction. $\endgroup$
    – OAH
    Oct 2, 2013 at 1:14
  • $\begingroup$ I added an expression for the moment generating function. $\endgroup$ Oct 2, 2013 at 2:17
  • $\begingroup$ I will check that. Thank you very much :) $\endgroup$
    – OAH
    Oct 3, 2013 at 4:53

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