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I am reading a paper and to strengthen my understanding I would like to know why $Z_{n} = \sum_{k=0}^{n} \frac{\Delta_{k}}{P(N\geq k)}\mathbf{1}_{N\geq k}$ converges almost surely to $Z= \sum_{k=0}^{N}\frac{\Delta_k}{P(N\geq k)}$, where $\Delta_{n} = X_n-X_{n-1}$.

It is given that $X_{n} \rightarrow X $ in $L^2$ $\Rightarrow \mathbf{E}(X_n) \rightarrow \mathbf{E}(X)$ as $n \rightarrow \infty$. N is a finite and nonnegative integer-valued random variable such that $P(N\geq n) > 0$ for $n \geq 0 $.

I do know the definition of almost sure convergence but quite frankly don't know how to apply it.

Help is much appreciated.

Maths student G
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  • For each given $\omega$, note that $N(\omega)$ is a finite, non-negative integer. So if we write $Y_k=\Delta_k/P(N\geq k)$ for simplicity, then $$Z_n(\omega)=\sum_{k=0}^{n}Y_k(\omega)\mathbf{1}{N(\omega)\geq k}=\sum{k=0}^{n\wedge N(\omega)}Y_k(\omega)$$ converges as $n\to\infty$ since this sequence eventually becomes constant with the value $\sum_{k=0}^{N(\omega)}Y_k(\omega)$. – Sangchul Lee Dec 12 '20 at 11:10
  • @Sangchul Lee thank you!:) I noticed that the sequence becomes constant but the way you wrote it makes it even clearer. Thanks again – Maths student G Dec 12 '20 at 11:18

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