Which of the following formulas will compute the variance of a discrete random variable X?
A)$\sum_{j=1}^{n}\left(x_{j}-E[X]^{2}\right)$

B)$\sum_{j=1}^{\infty} \mathrm{E}[X]^{2} f\left(x_{j}\right)$

C)$\sum_{j=1}^{\infty}\left(x_{j}-E[X]\right)^{2} f\left(x_{j}\right)$

D)$\sum_{j=1}^{\infty}\left(x_{j}\right) 2 f\left(x_{j}\right)$

Question

Which of the following formulas will compute the variance of a discrete random variable X?
A)$\sum_{j=1}^{n}\left(x_{j}-E[X]^{2}\right)$

B)$\sum_{j=1}^{\infty} \mathrm{E}[X]^{2} f\left(x_{j}\right)$

C)$\sum_{j=1}^{\infty}\left(x_{j}-E[X]\right)^{2} f\left(x_{j}\right)$

D)$\sum_{j=1}^{\infty}\left(x_{j}\right) 2 f\left(x_{j}\right)$

Quizplus · Accepted Answer

The formula for the variance of a discrete random variable X is $\operatorname{Var}(X)=\sum_{j=1}^{\infty}\left(x_{j}-E[X]\right)^{2} f\left(x_{j}\right)$. This is exactly option C. Option A has an error, since the squared difference should be with the expectation, rather than squared expectation. Option B only calculates the expectation of X squared, which is not the same as variance. Option D multiplies each value by its probability and does not have a term for calculating the squared difference from the expectation, so it is incorrect as well.

Which of the Following Formulas Will Compute the Variance of a Discrete