Lecture 10: Expectation Continued | Statistics 110
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers cite it as the source of the proof of linearity of expectation and of the lecture's total-probability argument that they are trying to understand.
Recommended by 0 people across 0 different months, 2016 to 2022.
What it is
Lecture 10 of Joe Blitzstein's Harvard Statistics 110 course. It proves linearity of expectation, works through a Putnam problem, introduces the Negative Binomial distribution, and ends with the St. Petersburg Paradox.
What people said
Cites Abhishek Bhatia · 2016-06 · Mathematics Stack Exchange · on “Proof of linearity for expectation given random variables are dependent”
The proof of linearity for expectation given random variables are independent is intuitive. What is the proof given there they are dependent? Formally, $$ E(X+Y)=E(X)+E(Y)$$ where $X$ and $Y$ are dependent random variables. The proof below assumes that $X$ and $Y$ belong to the sample space. That is, they map from the sample space to a real number line. Is that also a condition for linearity of expectation? Proof: $$E\left(X+Y\right) =\sum\limits_{s}\left(X+Y\right)\left(s\right) P\left({s}\right) $$ $$E\left(X+Y\right) =\sum\limits_{s}\left(X\left(s\right)+Y\left(s\right)\right) P\left({s}\right) $$ $$E\left(X+Y\right) =\sum\limits_{s} X\left(s\right)P\left({s}\right) + \sum\limits_{s} Y\le
Cites Anisha Jain · 2018-05 · Mathematics Stack Exchange · on “Law of Total Probability (Lecture 110 Harvard)”
I was watching Harvard STAT 110 lecture series on Youtube. As per lecture 10 [this video] , given $$T=X+Y$$ I have the following question: Why is $$P(T=t) = \sum P(T=t|X=x)\ P(X=x)$$ and not $$\sum P(T=t|X=x)\ P(X=x)+ \sum P(T=t|Y=y)\ P(Y=y)?$$
Cites Philemur · 2022-12 · Mathematics Stack Exchange · on “linearity of expectation, test scores and variance”
I got a very good answer to my last post [LINK] on the subject and traced the proof of linearity of expectation through this video [this video] . But I now have a total of three follow-up questions on the topic; sorry for that, but I have hardly done any maths for over a decade. If I have an MC test with 45 questions, then the expected value X of the test score is calculated as the sum of the expected values $E(Y_i)$ of all individual items for purely random answer patterns: $$E(X)=\sum_{i}E[Y_i]$$ Since there is one correct answer for each item and this yields exactly one point, the individual expected values of the item scores correspond to the reciprocal value of the amount of answer alternati