# Thread: moments of the sum of inverse ratio distribution

1. ## moments of the sum of inverse ratio distribution

I've had problems working this out:

Let $\displaystyle z_1,...,z_N$ be N independent uniform distributed variables that is $\displaystyle z_1,...,z_N$ are all iid uniform[0,1].

Define: $\displaystyle s_i = z_i/(z_1+...+z_N)$ for all i=1,...,N

That is, $\displaystyle s_i$ is the ratio of one of the variables divided by the sum of all variables, so that $\displaystyle s_1+...+s_N = 1$

I'm looking for the
- expected value: $\displaystyle E[s_i]$. I'm 99% sure this is equal to 1/N.

- variance: $\displaystyle VAR[s_i]$

- covariance: $\displaystyle COV[s_i,s_j]$

I would be very grateful for all help!

2. Originally Posted by Per
I've had problems working this out:

Let $\displaystyle z_1,...,z_N$ be N independent uniform distributed variables that is $\displaystyle z_1,...,z_N$ are all iid uniform[0,1].

Define: $\displaystyle s_i = z_i/(z_1+...+z_N)$ for all i=1,...,N

That is, $\displaystyle s_i$ is the ratio of one of the variables divided by the sum of all variables, so that $\displaystyle s_1+...+s_N = 1$

I'm looking for the
- expected value: $\displaystyle E[s_i]$. I'm 99% sure this is equal to 1/N.

- variance: $\displaystyle VAR[s_i]$

- covariance: $\displaystyle COV[s_i,s_j]$

I would be very grateful for all help!
$\displaystyle E[s_i] = \frac{\frac{1}{2}}{\frac{1}{2}N}$ Now, your 100% sure. []Edit[]Oops.

As for variance, try obtaining the second moment.

$\displaystyle E[s_i^2] = E[\frac{z_i ^2}{(z_1 +... + z_N)^2}]$

Then you can find the variance and covariance from the definition.

3. Originally Posted by Per
I've had problems working this out:

Let $\displaystyle z_1,...,z_N$ be N independent uniform distributed variables that is $\displaystyle z_1,...,z_N$ are all iid uniform[0,1].

Define: $\displaystyle s_i = z_i/(z_1+...+z_N)$ for all i=1,...,N

That is, $\displaystyle s_i$ is the ratio of one of the variables divided by the sum of all variables, so that $\displaystyle s_1+...+s_N = 1$

I'm looking for the
- expected value: $\displaystyle E[s_i]$. I'm 99% sure this is equal to 1/N.
Let me add the final percent: you have $\displaystyle 1=E[s_1+\cdots+s_N]=E[s_1]+\cdots+E[s_N]=NE[s_i]$ by symmetry hence $\displaystyle E[s_i]=1/N$
. (To Anonymous: you can't compute the expectation like you did!)

- variance: $\displaystyle VAR[s_i]$
So you need $\displaystyle E[(s_1)^2]$. I'm not sure you can't avoid computing the $\displaystyle N$-fold integral $\displaystyle \int_0^1\cdots\int_0^1\frac{(x_1)^2}{(x_1+\cdots+x _N)^2}dx_1\cdots dx_N$. Did you already try?

- covariance: $\displaystyle COV[s_i,s_j]$
This one can follow from the previous one, in a similar way to $\displaystyle E[s_1]$.

4. I thought of doing the N-fold integral but wanted to convince myself that there isn't any easier way of doing it.