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
2 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.
The second term is such because $E (X)$ is a constant, and the expectation of a constant is the constant itself (same for the last term ($E (X))^2$) $=E (X^2)-2 (E (X))^2+ (E (X))^2=E (X^2)- (E (X))^2$
Difference between logarithm of an expectation value and expectation value of a logarithm Ask Question Asked 14 years, 10 months ago Modified 10 years, 10 months ago
How did you reformulate the density function into an arithmetico-geometric series? Otherwise, the result for the sum of geometric series is very well known and it is not unusual to transform a mathematical expression into another, familiar one via simple operations to make the solution apparent. It doesn't mean your approach is wrong, but you might have needed to check Wikipedia to get the result.
The expected value of a function can be found by integrating the product of the function with the probability density function (PDF). What if I want to find the expected value of the PDF itself? Th...
I think you mean the expectation of the square of a geometric random variable. "Expectation squared ____" would seem to mean "square of expectation ____".
