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I found the following in He, Wang, Yan's "Semimartingales":

It is well known that for any non-negative random variable $X$ one can define conditional expectation by $\mathbb E\left[X\mid\mathcal G\right]=\lim_\infty \mathbb E\left[X\wedge n\mid\mathcal G\right]$. However, even if $X$ takes finite values, $\mathbb E\left[X\mid\mathcal G\right]$ may be $+\infty$ on a set with positive probability.

I wonder why it is true since conditional expectation is an "average", if $X$ take finite values,so is the average. Can anyone give my a counterexample?

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  • $\begingroup$ Hint: Develop completely the case when $\mathcal G$ is finite. ("if $X$ (almost surely) take(s) finite values,so is the average" No.) $\endgroup$ – Did Sep 26 '15 at 15:07
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If the question is how an expectation of a random variable that only takes finite values can be infinite, one of the most well known examples is this: $$ X =2^n\text{ with probability } \frac 1 {2^n} \text{ for }n=1,2,3,\ldots $$ This just means every time you toss the coin your winnings double. Toss the penny. If you get "heads", you win $1$ cent. If tails, toss again and if you get heads you win $2$ cents. If tails the first two times, toss a third time and if you win you get $4$ cents. And then $8$, and so on. Would you pay $\$100$ for each coin toss to play this? If so, you'll ultimately come out ahead, because some day you'll hit one of those rare instances where you get so many consecutive tails that your winnings will exceed all your vast losses up to that point. And the same is true if you pay $\$1$ trillion each time. Now matter how big the amount you pay each time, you ultimately come out ahead.

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