Computing distributions by using convolution.

MathematicalPhysicist

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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:
[tex]f_U(u)=\int_{-\infty}^{\infty}f_X(t)f_Z(u-t)dt[/tex]
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 doesnt apply for a difference between random variables, any tips?
 

gel

532
4
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,

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

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

[tex]
E(f(U))=\int_{-1}^1 p(u)f(u)\,du
[/tex]

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

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:
[tex]f_U(u)=\int_{-\infty}^{\infty}f_X(t)f_Z(u-t)dt[/tex]
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 doesnt 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 [tex]Z=-Y[/tex] 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:

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

Okay, the first thing to notice is that the definitions of [tex]f_X[/tex] and [tex]f_Z[/tex] 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 [tex]0<t<1[/tex] and [tex]-1<u-t<0[/tex]. Solving the second expression for [tex]t[/tex] gives us [tex]u < t < u+1 [/tex]. Notice that this is a function of [tex]u[/tex], which is an independent variable. Thus, we're going to get a piecewise expression in for the answer, in terms of [tex]u[/tex]:

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

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 [tex]f_X(t)[/tex] and [tex]f_Z(u-t)[/tex], and estimate the area under their product. Then, try it for a few different values of [tex]u[/tex], and you should see a pattern emerging.
 

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