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Given the sequence $(X_n), n=1,2,... $, of iid exponential random variables with parameter $1$, define:

$$ M_n := \max \left\{ X_1, \frac{X_1+X_2}{2}, ...,\frac{X_1+\dots+X_n}{n} \right\} $$ I want to calculate $\mathbb{E}(M_n)$.

Define for $k = 1, 2, ..., n$

$$Y_k := \frac{X_1 + ... + X_k}{k}$$

Then we have $E[Y_k] = 1$

We can rewrite:

$$M_n = \sum_{k=1}^{n} Y_k 1_{A_{n,k}}$$

where $A_{n,k}$ is the event that $Y_k = \max\{Y_1, ..., Y_n\}$

I was thinking that $E[M_n] = 1$ using independence, but then again, intuitively, the larger $Y_k$ is, the greater is the probability that it is the maximum.

Why rigorously is it that $\sigma(Y_k)$ and $\sigma(A_{n,k})$ are not independent?

I know that $\sigma(Y_k) = \{Y_k^{-1}(B) | B \in \mathscr B\}$ but have no clue as to what Borel set to use.

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    $\begingroup$ The terminology may be misleading you. The definition of independence of sigma algebras is actually about the probability measures defined on them. Therefore, you can prove lack of independence by choosing suitable events and doing a probability calculation. $\endgroup$
    – whuber
    Dec 18, 2015 at 18:31
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    $\begingroup$ @BCLC All the $Y_k$ have mean one, whether or not they're independent (they aren't) doesn't change this. I think pursuing the question of independence here won't help you calculate $E(M_n)$. $\endgroup$
    – dsaxton
    Dec 18, 2015 at 19:09
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    $\begingroup$ Fleshing out @whuber's comment: the $\sigma$-algebras $\mathcal{F}$ and $\mathcal{G}$ are independent iff all pairs of events $F \in \mathcal{F}$ and $G \in \mathcal{G}$ are independent. Thus you may prove the $\sigma$-algebras in question are not independent by finding an event $F$ that depends only $Y_k$ but $\mathbb{P}(F,A_{n,k}) \neq \mathbb{P}(F)\mathbb{P}(A_{n,k})$. dsaxton essentially gave you an example :) $\endgroup$ Dec 18, 2015 at 19:09
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    $\begingroup$ Btw, I commend your effort on (apparently) working through a probability text book :) $\endgroup$ Dec 18, 2015 at 19:51
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    $\begingroup$ @BCLC Mathematical rigor aside (I don't see why it's needed here), it should be fairly intuitive why these are not independent. If you're told that $Y_1 < \epsilon$ is $P(A_{n, 1})$ left unchanged? What if $\epsilon$ is made small and $n$ large? $\endgroup$
    – dsaxton
    Dec 18, 2015 at 20:01

1 Answer 1

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Consider the easiest non-trivial case of $n=2$, $k=1$.

Thus, $Y_k = Y_1 = X_1$ and $Y_2 = (X_1 + X_2)/2$.

Observe that $A_{n,k} = A_{2,1} = \{Y_1 \geq Y_2\} = \{X_1 \geq X_2\}$, so we have $\mathbb{P}(A_{n,k}) = 1/2$ (since the $X_i$'s are iid).

Now examine the event that $Y_1 < t$ (any fixed $t > 0$). This has probability $\mathbb{P}(Y_1 < t) = 1-e^{-t}$.

On the other hand, $$ \mathbb{P}(Y_1 < t, A_{2,1}) = \mathbb{P}(Y_1 < t, Y_1 \geq Y_2) = \mathbb{P}(X_2 \leq X_1 < t). $$

The rhs can be calculated as $$ \int_0^t \mathbb{P}(X_2 \leq u)e^{-u}du = \int_0^t (1 - e^{-u})e^{-u}du $$ which is very much not equal to $\frac{1}{2}(1-e^{-t})$!

[Edit: n.b. this answer concerns the independence, not how to calculate the expected maximum.]

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  • $\begingroup$ Ah, if only I remembered $\sigma(Y_k) = \{Y_k \in (\infty, -a] | a \in \mathbb R\}$. Thanks P.Windridge ^-^ $\endgroup$
    – BCLC
    Dec 18, 2015 at 20:03
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    $\begingroup$ No probs :) Rather than beginning from "find a Borel set such that [blah]" it's more helpful to try imagining what's going on intuitively, and trying to make it rigorous afterwards (if necessary) as @dsaxton suggests. This is one of the basic idea in Polya's How to Prove it book. $\endgroup$ Dec 18, 2015 at 20:10
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    $\begingroup$ (+1) BCLC: Since this question really concerns definitions, it seems worthwhile to nit-pick: $\sigma(Y_k)$ is much larger than just that collection of semi-infinite intervals! It is only generated by them. It includes much, much more complicated sets. This gives you far more flexibility than you otherwise might have thought in choosing an event to supply as a counterexample to the hypothesis of independence. $\endgroup$
    – whuber
    Dec 18, 2015 at 20:21
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    $\begingroup$ The formula in your comment is not entirely correct, as I have tried to explain. By definition, the countable union and finite intersection of elements of $\sigma(Y_k)$ must also be in $\sigma(Y_k)$, but that obviously is not the case for the collection you have written. Your expression has two problems. The first is that your intervals are all empty: you probably meant to write "$(-\infty, a]$." Second, that's still insufficient, because your collection doesn't contain (for instance) the interval $(0,1]=(-\infty,1] - (-\infty,0]$, which is in $\sigma(Y_k)$. $\endgroup$
    – whuber
    Dec 19, 2015 at 0:49
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    $\begingroup$ Yes, that conveys the right idea. The main thing to remember is that it gives you great flexibility to select measurable subsets when assessing independence (or the lack thereof). $\endgroup$
    – whuber
    Dec 19, 2015 at 20:23

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