# Why is Convolution Well-Defined (Simple Example)

I am having trouble understanding why convolution is well-defined.

Let's take a simple example:

$$(\Omega, \mathcal{F}, P)$$ probability space and $$X_{1}, X_{2}$$ two real random variables where $$P(X_{1}=3)=\frac{1}{2},P(X_{2}=2)=\frac{1}{4}$$

And $$P(X_{1}=1)=\frac{1}{5},P(X_{2}=4)=\frac{1}{3}$$

Then my understanding of convolution is

$$P_{X_{1}+X_{2}}\circ A^{-1}$$ where $$A: X_{1} \times X_{2}\to \mathbb R,A(x_{1},x_{2})=x_{1}+x_{2}$$

So surely, if, for instance $$X_{1}+X_{2}=5$$, I get more than one preimage, and hence how can convolution be well-defined?

In the above case, I would get:

$$P_{X_{1}+X_{2}}\circ A^{-1}(5)=P_{X_{1}}(3)P_{X_{2}}(2)=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$$ while

$$P_{X_{1}+X_{2}}\circ A^{-1}(5)=P_{X_{1}}(1)P_{X_{2}}(4)=\frac{1}{5}\times\frac{1}{3}=\frac{1}{15}$$

I do not know where I am going wrong in my understanding of convolution. Any help is greatly appreciated.

If $$P$$ denotes the probability measure on $$(\Omega,\mathcal F)$$ then it induces for every random variable $$Z$$ a probability measure $$P_Z$$ on $$(\mathbb R,\mathcal B)$$ that is prescribed by:$$B\mapsto P(\{\omega\in\Omega\mid Z(\omega)\in B\})=P(\{Z\in B\})=P(Z\in B)$$

Here $$\{Z\in B\}$$ abbreviates $$\{\omega\in\Omega\mid Z(\omega)\in B\}$$ and $$P(Z\in B)$$ abbreviates $$P(\{Z\in B\})$$

So we have $$P_Z(B)=P(\{Z\in B\}$$ for Borel subsets of $$\mathbb R$$.

Another notation of this probability is $$PZ^{-1}$$ prescribed by:$$B\mapsto P(Z^{-1}(B))=P(\{\omega\in\Omega\mid Z(\omega)\in B\})=P(\{Z\in B\})=P(Z\in B)$$

Observe that $$P_Z$$ and $$PZ^{-1}$$ are notations for the same measure.

Also random vector $$(X_1,X_2)$$ induces a probability measure.

This time denoted as $$P_{(X_1,X_2)}$$ and defined on $$(\mathbb R^2,\mathcal B^2)$$.

If $$A:\mathbb R^2\to\mathbb R$$ is prescribed by $$(x,y)\mapsto x+y$$ then $$A$$ is a Borel-measurable function.

That means that it can be looked at as a random variable on space $$(\mathbb R^2,\mathcal B^2,P_{(X_1,X_2)})$$.

Applying the principle that was mentioned above on space $$(\mathbb R^2,\mathcal B,P_{(X_1,X_2)})$$ we have measure $$P_{(X_1,X_2)}A^{-1}$$ on $$(\mathbb R,\mathcal B)$$ and it is not difficult to deduce that:$$P_{X_1+X_2}=P_{A\circ(X_1,X_2)}=P_{(X_1,X_2)}A^{-1}$$

In your question you mix up the two notations, and this can be a source of confusion on its own.

If $$X_1,X_2$$ are random variables then so is $$X_1+X_2$$.

This with e.g.:$$P_{X_1+X_2}(\{5\})=P(X_1+X_2\in\{5\})=P(X_1+X_2=5)$$

If moreover $$X_1,X_2$$ only take integers as value then this can be expanded to:$$\cdots=\sum_{n,m\in\mathbb Z\wedge n+m=5}P(X_1=n\wedge X_2=m)$$

• On your last point, if $X_{1}, X_{2}$ are considered independent random variables, can I say $\sum_{n,m \in \mathbb Z, n + m =5}P(X_{1}=n)P(X_{2}=m)$ Jan 28, 2019 at 17:51
• Yes, that is correct. Jan 28, 2019 at 18:57