# Computing distributions by using convolution.

• MathematicalPhysicist
This is the basis for the convolution calculation.In summary, we use the convolution formula to find the distribution of U=X-Y, where X and Y are independent and uniformly distributed on [0,1]. This leads to a triangular distribution for U, rather than a uniform distribution as initially thought. We can also use graphical methods to better understand this calculation.
MathematicalPhysicist
Gold Member
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?

loop quantum gravity said:
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,

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

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

$$E(f(U))=\int_{-1}^1 p(u)f(u)\,du$$

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

loop quantum gravity said:
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.

loop quantum gravity said:
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 $$0<t<1$$ and $$-1<u-t<0$$. Solving the second expression for $$t$$ gives us $$u < t < u+1$$. 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 $$u$$:

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

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.

## 1. What is convolution and why is it used in computing distributions?

Convolution is a mathematical operation that combines two functions to produce a third function. It is used in computing distributions because it allows for the calculation of the probability of a particular outcome by combining the probabilities of all possible outcomes.

## 2. How does convolution work in computing distributions?

In computing distributions, convolution involves multiplying the probability density functions (PDFs) of two random variables and then integrating the result over all possible values of the variables. This produces the PDF of the sum of the two variables, which represents the distribution of the combined outcome.

## 3. What are the advantages of using convolution in computing distributions?

Using convolution allows for the simplification of complex distributions by breaking them down into smaller, more manageable parts. It also allows for the calculation of the distribution of the sum of multiple variables, which is useful in many real-world applications.

## 4. Are there any limitations to using convolution in computing distributions?

One limitation of using convolution is that it assumes that the variables being combined are independent of each other. In cases where this assumption does not hold, the resulting distribution may not accurately reflect the true distribution of the combined outcome.

## 5. Can convolution be used for any type of distribution?

Yes, convolution can be used for any type of distribution as long as the variables being combined are continuous and independent. This includes normal, exponential, and uniform distributions, among others.

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