$\sum_{k=0}^{\infty} k \textbf{1}_{ \{X_1 + X_2 =k \} } = X_1 + X_2$

Question

Could someone explain to me why the following holds ?

$X_1, X_2 - \textrm{poisson - distributed random variables}$

$\sum_{k=0}^{\infty} k \textbf{1}_{ \{X_1 + X_2 =k \} } = X_1 + X_2$

Answer 1

This statement is not true if some $\omega\in\Omega$ exists with $X_1(\omega)+X_2(\omega)\notin\{0,1,2,\dots\}$.

This because the LHS can only take values in $\{0,1,2,\dots\}$.

Let us assume that such $\omega$ does not exist.

Fix some $\omega\in\Omega$

Observe that for every index $k$ with $X_1(\omega)+X_2(\omega)\neq k$ we have $\mathbf1_{\{X_1+X_2=k\}}(\omega)=0$ and consequently $k\mathbf1_{\{X_1+X_2=k\}}(\omega)=0$.

So all terms in the summation vanish except the term $k\mathbf1_{\{X_1+X_2=k\}}(\omega)$ with $k=X_1(\omega)+X_2(\omega)\in\{0,1,2,\dots\}$.

Note that this term takes value $X_1(\omega)+X_2(\omega)$.

Proved is now that $\sum_{k=0}^{\infty}k\mathbf1_{\{X_1+X_2=k\}}(\omega)=X_1(\omega)+X_2(\omega)$.

According to our assumption this is true for every $\omega\in\Omega$ so we conclude that:$$\sum_{k=0}^{\infty}k\mathbf1_{\{X_1+X_2=k\}}=X_1+X_2$$

Answer 2

Ultimately, because $X_1$ and $X_2$ are (for some unstated reason) non-negative integer valued random variables. On the left you have a sum of terms all of which are 0 except for possibly the $k$th term, which is equal to $k$ when $X_1+X_2 =k$. On the right, the value of $X_1+X_2$, whatever it is. Since the sum ranges over non-negative integer $k$, the equality holds only when $X_1+X_2$ takes on such a value. In this case you could think of this as a case of the Sherlock Holmes principle: when you have eliminated the impossible, whatever remains, blah blah.