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Let $X$ and $Y$ be continuous random variables, while $N$ be a discrete random variable.

The math assistant said that 4 is the answer for the problem that $\mathbb{E}(X|Y)$ is

  1. a number.
  2. a discrete random variable.
  3. a continuous random variable.

  4. not determined due to the lack of information.

    • Since if $X$ and $Y$ are independent, $X$ is a number. Otherwise, $X$ is a continuous random variable.

I agree the above answer and the reason. However, I do not understand why the answer is 2 for the following problem that $\mathbb{E}(\mathbb{E}(X|Y,N)|N)$ is

  1. a number.
  2. a discrete random variable.
  3. a continuous random variable.
  4. not determined due to the lack of information.

Can anyone explain why?

Danny_Kim
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1 Answers1

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$E(X|Z)$ ls always of the form $f(Z)$ for some measurable function $f: \mathbb R \to \mathbb R$. Hence the conditional probability above is of the form $f(N)$ and this can take only countable number of values.

Kavi Rama Murthy
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    I am wondering if $/mathbb{E}(/mathbb{E}(X|Y,N)|N)$X$ is a number if $X$ is independent of both $Y$ and $N$, isn’t it?? – Danny_Kim May 12 '20 at 09:22
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    It could be a number. But the point is the conditional probability can take at most countably many values, so it is a discrete random variable. @Danny_Kim – Kavi Rama Murthy May 12 '20 at 09:25
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    If so, why the answer of the first question is not 2 but 4? – Danny_Kim May 12 '20 at 09:26
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    If you say that the answer is 2) you are saying that $E(X|Y)$ must be be discrete. There are examples where is is not, so 2) is not the right option for the first question. In the second question the conditional probability is necessarily discrete , so 2) is the right option. @Danny_Kim – Kavi Rama Murthy May 12 '20 at 09:39
  • I am really sorry. I mean number 3. Not 2. That is, as the same principle, the answer of the first question is 3. – Danny_Kim May 12 '20 at 09:40
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    It is already mentioned that in the first question $E(X|Y)$ can be a constant and it can also be a continuous random variable. So 2) and 3) are both false in the first question. – Kavi Rama Murthy May 12 '20 at 09:42