Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose I have $S=U_1+U_2+\dots+U_n$ where $U_i$ are distributed Uniform$[-1,1]$.

I am trying to show a couple of things. First, what is the characteristic function. I can show this easily enough for a single uniform RV, but for some reason the sum is not coming to me.

Second, does the inversion formula apply as long as $n\ge2$? and if so, how can I show this?

Finally, how can I use the inversion formula to show a real-valued integral for the density (i.e getting rid of the imaginary part of the characteristic function)?

I have worked through how to derive a single $U$, but these extensions escape me.

Many thanks for the help~

share|improve this question
    
Are your $U_j$ independent? One of the most important properties of characteristic functions is about the characteristic function of the sum of independent random variables. –  Robert Israel Aug 13 '12 at 16:37
    
yes they are iid....and I believe that means that I can take the product of their individual characteristic functions, correct? I'm still somewhat fuzzy on the second and third parts though. –  Justin Aug 13 '12 at 16:39
    
38 minutes. $ $ –  Did Aug 13 '12 at 18:54
    
@did, what are you talking about? –  Justin Aug 13 '12 at 18:57
    
This is the time-span between the moment when you posted your question and the moment when you accepted an answer. Such a short time-span has disadvantages, some (or all) of which you can surely guess. –  Did Aug 13 '12 at 19:08

1 Answer 1

up vote 4 down vote accepted

The characteristic function is an expectation: $$ \varphi_S(t) = \mathbb{E}\left(\exp(i S t)\right) = \mathbb{E}\left(\exp\left(i \left(U_1 + U_2 + \cdots + U_n \right) t\right)\right) $$ Now, if $U_i$ is independent, the expectation factors into product of expectations, because : $$ \varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot \mathrm{e}^{i t U_2} \cdots \mathrm{e}^{i t U_n}\right) = \mathbb{E}\left(\mathrm{e}^{i t U_1} \right)\cdot \mathbb{E}\left(\mathrm{e}^{i t U_2} \right) \cdots \mathbb{E}\left(\mathrm{e}^{i t U_n}\right) $$ Since $U_i$ have identical distributions, these expectations are the same, thus: $$ \varphi_S(t) = \left(\mathbb{E}\left(\mathrm{e}^{i t U} \right)\right)^n = \left( \frac{\sin(t)}{t}\right)^n $$

According to the inversion formula, there is one-to-one correspondence between cumulative distribution function, and the characteristic function. Since $\varphi_S(t)$ is a characteristic function for all $n \in \mathbb{N}$ by construction, the inversion formula applies for all natural $n$.

Moverover, if $\varphi_S(t)$ is integrable, then $F_S(x)$ is absolutely continuous, i.e. the notion of the probability density function is well-defined. Notice that $\varphi_S(t)$ is absolutely integrable for $n \geqslant 2$, by virtue of $|\varphi_S(t)| \leqslant \min\left(1, t^{-n}\right)$, and hence integrable. However, $\varphi_S(t)$ is also conditionally integrable for $n=1$ (the integral is known as Dirichlet integral).

Applying the inversion formula to obtain an explicit expression for the probability density is somewhat involved. This earlier post of mine contains the derivation.

To conclude, let me remark, that the sum of iid uniform continuous random variable follows what is known as the Irwin-Hall distribution.

share|improve this answer
    
Looks like for Irwin-Hall the U$_i$ are all U[0,1] and not U[-1,1]. –  Michael Chernick Aug 13 '12 at 17:31
1  
Yes, but the rescaling is trivial. –  Sasha Aug 13 '12 at 17:32
    
@Sasha, thank you for the help! Speaking in general, if I want to derive the moments of a distribution using its characteristic function, do I just take the "nth" derivative with respect to the parameter and evaluate at t=0? If so, how does this apply to a distribution with multiple parameters (such as Gamma($\alpha,\Beta$)? –  Justin Aug 13 '12 at 17:35
    
@Sasha Okay but be careful with your terminology. –  Michael Chernick Aug 13 '12 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.