Computing distributions by using convolution.

[MathematicalPhysicist]

Let X,Y~U(0,1) independent (which means that they are distributed uniformly on [0,1]). find the distribution of U=X-Y.
well intuitively U~U(-1,1), but how to calculate it using convolution.
I mean the densities are f_Z(z)=1 for z in [-1,0] where Z=-Y and f_X(x)=1 for x in [0,1], now i want to calculate using convolution i.e:
f_U(u)=\int_{-\infty}^{\infty}f_X(t)f_Z(u-t)dt
where t in [0,1] and u-t in [-1,0] so u is in [-1,1], as i said i know what intuitively it should be but i want to formally calculate it, i.e compute the integral, and t is between [u,u+1], but i think that this integral doesn't apply for a difference between random variables, any tips?

 

[gel]

intuitively U~U(-1,1)

No, that's not right. The density will be peaked at 0 - intuitively (to me), you would expect U to be more likely to be close to 0 than with the uniform density.

Shouldn't be too difficult to do this from first principles. For any function f:[-1,1]->R,

<br /> E(f(U))=E(f(X-Y))=\int_{0}^1\int_0^1 f(x-y)\,dxdy \textrm{ (by independence).}<br />

Rearrange this integral by the change of variables u=x-y on the inner integral to get something like

<br /> E(f(U))=\int_{-1}^1 p(u)f(u)\,du<br />

and then p(u) will be the density you're looking for.

 

[quadraphonics]

Let X,Y~U(0,1) independent (which means that they are distributed uniformly on [0,1]). find the distribution of U=X-Y.
well intuitively U~U(-1,1), but how to calculate it using convolution.

No, it's definitely *not* uniform. The Central Limit Theorem, for example, tells you that it can't be.

I mean the densities are f_Z(z)=1 for z in [-1,0] where Z=-Y and f_X(x)=1 for x in [0,1], now i want to calculate using convolution i.e:
f_U(u)=\int_{-\infty}^{\infty}f_X(t)f_Z(u-t)dt
where t in [0,1] and u-t in [-1,0] so u is in [-1,1], as i said i know what intuitively it should be but i want to formally calculate it, i.e compute the integral, and t is between [u,u+1], but i think that this integral doesn't apply for a difference between random variables, any tips?

It's not a problem that the integral doesn't apply to differences, as the substitution Z=-Y has changed the problem to a sum of (independent) random variables. Rather, the confusion is probably because you're assuming a wrong result (uniform on [-1,1]), instead of the correct one (a triangular distribution). Anyhow, let's proceed with the integral and see what happens:

f_U(u) = \int_{-\infty}^{\infty} f_X(t)f_Z(u-t)dt

Okay, the first thing to notice is that the definitions of f_X and f_Z are piece-wise, and so we'll need to consider all the relevant cases and write the integral in a piecewise manner. First, we need to identify the region of integration where both terms in the integrand are non-zero (everything else we can ignore). This requires both 0&lt;t&lt;1 and -1&lt;u-t&lt;0. Solving the second expression for t gives us u &lt; t &lt; u+1. Notice that this is a function of u, which is an independent variable. Thus, we're going to get a piecewise expression in for the answer, in terms of u:

f_U(u) = \left\{ \begin{array}{l} <br /> 0\; \mathrm{if}\, u &lt; -1 \\ <br /> \int_0^{1+u}dt\; \mathrm{if}\, -1 \leq u &lt; 0 \\<br /> \int_u^1dt\; \mathrm{if}\, 0 \leq u &lt; 1\\<br /> 0\; \mathrm{if}\, u \geq 1\\<br /> \end{array}\right.

Now, all that remains is to evaluate the two easy integals and observe that it is the so-called triangular distribution.

This particular problem (convolution of uniform densities) is handily demonstrated graphically. Try picking a particular u, drawing f_X(t) and f_Z(u-t), and estimate the area under their product. Then, try it for a few different values of u, and you should see a pattern emerging.