$\mathbb{E}[S_n]$, when $S_n$ takes $X_1+...+X_n$ or $Y_1+...+Y_n$?

I don't think it is the same thing as:

Why does $\mathbb{E}(\frac{X_1+...+X_n}{n})=\mathbb{E}(X)$?

because the former involves conditional expectation.

  • 2
    $\begingroup$ I agree it's not a duplicate. When it was originally posted, it appeared to focus on the question the community identified as a duplicate. After it was edited, it more clearly focused on iterated expectation. As it stands, it does appear to duplicate other threads about iterated expectations, such as stats.stackexchange.com/questions/69399. Should we reopen the question and close it as a duplicate of that one? $\endgroup$
    – whuber Mod
    Dec 23, 2015 at 15:17

1 Answer 1


I don't think it is quite a duplicate.

Of course as soon as you apply the law of total expectation to the 50/50 part the expectation of the components are then each answered by that other question.

[One might argue that it's a duplicate simply by applying linearity of expectation twice, but I think that's perhaps a bit indirect for a duplicate.]


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