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.


1 Answer 1


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:$$\cdots=\sum_{n,m\in\mathbb Z\wedge n+m=5}P(X_1=n\wedge X_2=m)$$

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

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