The expectation of an expectation - Mathematics Stack Exchange This may seem trivial but just to confirm, as the expected value is a constant, this implies that the expectation of an expectation is just itself It would be useful to know if this assumption is correct or if any subtleties cause this not to be true i e $ \mathbb{E[\mathbb{E[x]}]}=\mathbb{E[x]} $
Calculate expectation of a geometric random variable A clever solution to find the expected value of a geometric r v is those employed in this video lecture of the MITx course "Introduction to Probability: Part 1 - The Fundamentals" (by the way, an extremely enjoyable course) and based on (a) the memoryless property of the geometric r v and (b) the total expectation theorem
Expectation of Minimum of $n$ i. i. d. uniform random variables. $\begingroup$ There's no sense in taking the derivative and then integrating; you can just use the tail-sum formula to calculate the expectation directly from the integral $\endgroup$ – Zaz Commented Apr 25, 2022 at 23:36
What is the difference between Average and Expected value? The distinction is subtle but important: The average value is a statistical generalization of multiple occurrences of an event (such as the mean time you waited at the checkout the last 10 times you went shopping, or indeed the mean time you will wait at the checkout the next 10 times you go shopping)
Expected value of a Gaussian - Mathematics Stack Exchange $\begingroup$ You're very welcome And you got it! The integral over $(-\infty, \infty)$ of a pdf results in 1 (which intuitively makes perfect sense, since when you integrate a pdf over an interval, you are calculating the probability that the random variable lands in that interval, so when you integrate a pdf over $(-\infty, \infty)$ you are calculating the probability that the random
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Difference between logarithm of an expectation value and expectation . . . To add on Didier's answer, it is instructive to note that the inequality ${\rm E}(\ln X) \le \ln {\rm E}(X)$ can be seen as a consequence of the AM-GM inequality combined with the strong law of large numbers, upon writing the AM-GM inequality $$ \sqrt[n]{{X_1 \cdots X_n }} \le \frac{{X_1 + \cdots + X_n }}{n} $$ as $$ \exp \bigg(\frac{{\ln X_1 + \cdots + \ln X_n }}{n}\bigg) \le \frac{{X_1