Talk:Wald's equation

Link is broken
The reference to encyclopedia of mathematics is broken. — Preceding unsigned comment added by Jakub Tětek (talk • contribs) 17:09, 8 July 2020 (UTC)

Untitled
Each X_i must be independent of N, or even just uncorrelated with N; they need not be i.i.d., they only need to have the same expectation. --Drew

It looks like the $$X_i$$ do not even have to be uncorrelated with $$N$$, it is sufficient if the event $$N=n$$ does not depend on $$X_{n+1}, X_{n+2}$$ etc, this is what is meant by the statement "$$N$$ can be a stopping time for the stochastic process $$X_i$$"; as mentioned here, for example. And somebody should check out Ross (1970), "Applied Probability Models with Optimization Applications", p.38 or so; I've seen a few references to it, but I don't have access to the book. &mdash;MoA)gnome (talk) 22:24, 22 March 2008 (UTC)

I agree - $$T$$ must be a stopping time, otherwise the identity is false... —Preceding unsigned comment added by 193.136.196.114 (talk) 12:23, 23 July 2008 (UTC)

I think the statements can be optimized somehow: on the one hand, by assuming independence between N and the X_n the basic version is essentially trivial (one needs not invoke Wald's equation in this case, and hence the reader is unlikely to have come to this page for this special case). On the other hand, assumption (2) is somewhat obscure. It is the same as N being a stopping time? If it is more general than that, one should make it clear somewhere that the theorem works with N being a stopping time, arguably one of the most common scenarios where the lemma is being used. I have added a comment in the discussion of the assumptions. — Preceding unsigned comment added by Queueing (talk • contribs) 13:11, 31 March 2015 (UTC)

Second proof is false
The second proof is false. In introduction of the article, it is stated that T can be a stopping time, yet proof #2 assumes that Xi is independent of T. However, Wald's identity holds for true stopping times, that is, T is independent only from the future (from XT+1, XT+2, ...), but can (and usually does) depend on X1, ... XT. —Preceding unsigned comment added by 193.77.126.73 (talk) 17:20, 15 May 2010 (UTC)

You were absolutely right. I corrected the text, clarifying that the second proof is only meant to prove a weaker statement (under stronger conditions). Szepi (talk) 00:41, 22 July 2010 (UTC)

Clarifications
Two things were spotted that need clarification "integrable random variable", "natural number" ... these are techical terms that a general reader cannot be expected to know. The problem might well be solvable by providing appropriate wikilinks, but this wasn't done. Someone with knowledge of exactly what is meant could probably find the right article/subsection to which to link. It was obvious what needed clarifying, so there was no need to spell it out here. Melcombe (talk) 10:03, 27 October 2010 (UTC)

Not very clear
I think that putting numbers from 1 to 12 to properties, and assuming some of them and citing them with their numbers makes the reading of this page unpleasant and more complicated that it should be. — Preceding unsigned comment added by 85.168.223.228 (talk) 22:18, 28 November 2012 (UTC)

Incorrect justification
The main article, in the section "Example where the number of terms depends on the sequence", says:
 * "Let $(X_{n})_{n∈$\mathbb{N}$}$ be a sequence of independent, symmetric, and ${–1, +1}$-valued random variables. For every $n ∈ $\mathbb{N}$$ let $\mathcal{F}_{n}$ be the σ-algebra generated by $X_{1},. . ., X_{n}$ and define $N = n$ when $X_{n}$ is the first random variable taking the value $+1$. Note that $P(N = n) = 1/2^{n}$, hence $E [ N ] < ∞$ by the ratio test."
 * My issue is that the ratio test correctly gives the finite value of the sum, but the expectation is not an infinite series because there could be a nonzero chance that $N$ will be infinite. You need to justify that N is integrable before you can use this reasoning. Lavaka (talk) 14:58, 23 November 2022 (UTC)