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Probability Tutorials

They must mean point-wise $\inf$ right? $g_n(w) := \inf_{k\geq n} f_k(w)$?

Thanks.

2 Answers2

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You pick some $w$ and find out what is $$\inf\{f_{n}(w),f_{n+1}(w),\ldots\}$$

That is the value of $g_n(w)$.

Pedro
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The PDF actually just defines $g_n=\inf_{k\ge n}f_k$, but yes, that means that for each $\omega\in\Omega$, $g_n(\omega)=\inf_{k\ge n}f_k(\omega)$, which is the pointwise infimum of $\{f_k:k\ge n\}$.

Brian M. Scott
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