How should one interpret an undefined expectation?

[ultrafilter]

So suppose we’ve got a Cauchy random variable, with density function f(x) = 1/(pi*(x[sup]2[/sup]+1)). The expectation of this variable is undefined, as the integral of x/(pi*(x[sup]2[/sup]+1)) over the real line does not converge. How should this be interpreted? I’ve always taken it to mean that no matter how large a reading I get on this variable, I can expect to see a larger reading in a finite number of trials. Is this correct, or even close to justifiable?

[IzzyR]

In essence this means that the expectation is infinite, or meaningless. (See also this thread.

[ultrafilter]

*Originally posted by IzzyR *
**In essence this means that the expectation is infinite, or meaningless. (See also this thread.

**

So if the expectation is infinite, is my interpretation of always being able to get a larger reading correct?

[IzzyR]

Truth is that I’m not so up on technical math lingo. I don’t see how you’d get an initial “reading”. But I would say that the value is larger than any conceivable value that you might assign to it.

[ultrafilter]

*Originally posted by IzzyR *
**Truth is that I’m not so up on technical math lingo. I don’t see how you’d get an initial “reading”. But I would say that the value is larger than any conceivable value that you might assign to it. **

According to my probability textbook (Jim Pitman’s Probability), the quotient of two standard normal variables has a Cauchy distribution. So there’s how I get an initial reading, and all subsequent readings.

The reason I ask is due to the Petersburg paradox, which I saw tonight on this page. There’s a good description of it there, so I’m just gonna cut and paste:

This is a paradox in the field of game theory. Consider a game in which a coin is flipped until heads comes up. If it comes up on the first toss, you win $1. If it comes up on the second toss, you win $2. If it comes up on the third toss, you win $4. In general, if it comes up on the nth toss, you win $2 n-1. The expectation (average amount you can expect to win) for this game can be found by adding together the products of the probabilities for each outcome with the amount that is won for each outcome. This gives us (1/2) x 1 + (1/4) x 2 + (1/8) x 4 + (1/16) x 8 + . . . = ½ + ½ + ½ + ½ + . . .

Since the number of terms in this sequence is infinite, the sum is also infinite. Therefore, the expectation for this game is infinite. The problem is that it doesn’t make sense to have an infinite expectation, especially since you will only win a finite amount of money everytime you play the game.

I think that my interpretation fits there, so I wondered if it was more generally applicable.

[Wumpus]

Slight hijack: the resolution of the St. Petersburg “paradox” is based in reality, not math. In other words, there are always real-world constraints not mentioned in the problem. Everyone knows that no casino has deep enough pockets to pay off on such a bet if the coin came up tails for a while. Since every real-world casino can only afford to lose so much, any similar bet offered in the real world would have a finite expectation.

Lewis Carroll offered a variant of the Petersburg paradox where a debtor managed to perpetually put off paying his debt by offering the bill collector twice as much if they collected it the next day. The next day, he did the same thing, and so on…

Just try that one on the bank.

[IzzyR]

*Originally posted by ultrafilter *

The problem is that it doesn’t make sense to have an infinite expectation, especially since you will only win a finite amount of money everytime you play the game.

I don’t understand the “problem”. The infinite expectation is a result of the infinite amount of coin flips that are being accounted for, not the amount that is won with each flip. Am I missing something?

[ultrafilter]

*Originally posted by IzzyR *
**

*Originally posted by ultrafilter *

The problem is that it doesn’t make sense to have an infinite expectation, especially since you will only win a finite amount of money everytime you play the game.

I don’t understand the “problem”. The infinite expectation is a result of the infinite amount of coin flips that are being accounted for, not the amount that is won with each flip. Am I missing something? **

Probably not. Your resolution makes sense.

[Race_Bannon]

*Originally posted by ultrafilter *
So suppose we’ve got a Cauchy random variable, with density function f(x) = 1/(pi(x[sup]2[/sup]+1)). The expectation of this variable is undefined, as the integral of x/(pi(x[sup]2[/sup]+1)) over the real line does not converge. How should this be interpreted? I’ve always taken it to mean that no matter how large a reading I get on this variable, I can expect to see a larger reading in a finite number of trials. Is this correct, or even close to justifiable? **

I’m not sure your “expect to see a larger reading…” statement is true, but I think I see where you’re going.

The interpretation I’ve had (of the expected value being undefined) is that for any finite “sample mean” you cannot expect the sample mean to converge to anything as the number of samples increases. It just drifts around. This is because the probability of getting an “outlier” is sufficient enough to jerk the sample mean anywhere at any time.

Check out this cite from a class I took for a nifty applet that lets you play around with this (see example 17).

Check it out at least 0.99999…ce

[ultrafilter]

*Originally posted by kellymccauley *
The interpretation I’ve had (of the expected value being undefined) is that for any finite “sample mean” you cannot expect the sample mean to converge to anything as the number of samples increases. It just drifts around. This is because the probability of getting an “outlier” is sufficient enough to jerk the sample mean anywhere at any time.

That makes a lot of sense. Until I hear otherwise, I’m going to assume that’s true.

**Check it out at least 0.99999…ce **

But that would take forever!