In My book I saw the following claim:
$E[X]=E[E[E[X|Y]|Z]]$
Is there a proof for this claim, as it was provided as is, with no further explanation...
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2Are you familiar with the tower property of conditional expectations? – Jose Avilez Dec 26 '21 at 22:12
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1This holds by definition of the conditional expectation ! – Surb Dec 26 '21 at 22:14
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I know that [[[|]|]]=[[[|]|=z]] for all z... – Dan Dec 26 '21 at 22:18
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1Do you know the rule $E[X]=E[E[X\mid Y]]$? If so, you can let $W = E[X\mid Y]$, so using that rule, we have $E[W]=E[E[W\mid Z]]=E[E[E[X\mid Y]\mid Z]]$. But $E[X]=E[W]$. So $$E[X]=E[E[E[X\mid Y]\mid Z]].$$ – Minus One-Twelfth Dec 26 '21 at 22:19
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@Surb My book only says E[X|Y]=E[X|Y=y] – Dan Dec 26 '21 at 22:19
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@Dan that is not correct – Jose Avilez Dec 26 '21 at 22:20
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@Dan: I would be very surprised ! $\mathbb E[X\mid Y]$ is a r.v. whereas $\mathbb E[X\mid Y=y]$ is not. – Surb Dec 26 '21 at 22:20
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1@JoseAvilez what isn't correct? – Dan Dec 26 '21 at 22:21
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@Dan $E[X , | , Y] \neq E[X , | , Y = y]$. See Surb's comment. – Jose Avilez Dec 26 '21 at 22:22
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1that's a well know law @Surb for calculating expected value we scan all possible values for y – Dan Dec 26 '21 at 22:42
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https://math.stackexchange.com/questions/41536/intuitive-explanation-of-the-tower-property-of-conditional-expectation – Shannon Starr Dec 27 '21 at 01:33
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You are essentially being asked to prove an iterated use of the law of total expectation. I suggest you start by looking up a proof of that law and then see if you can extend it to the iterated version you have here.
Ben
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