Variance and Expected ValueDate: 09/12/2004 at 19:57:58 From: Frank Subject: Variance is equal to the expected value How can you prove that the variance is equal to the expected value of the square of the random variable minus the mean squared? Date: 09/13/2004 at 09:32:23 From: Doctor Luis Subject: Re: Variance is equal to the expected value Hi Frank, For a random variable X, the mean mu is the expected value E[X], and the variance Var(X) is the expected value of (X - mu)^2. So, mu = E[X] Var(X) = E[(X - mu)^2] = E[X^2 - 2mu*X + mu^2] = E[X^2] - E[2mu*X] + E[mu^2] = E[X^2] - 2mu*E[X] + mu^2 = E[X^2] - 2mu*mu + mu^2 = E[X^2] - 2mu^2 + mu^2 = E[X^2] - mu^2 where I've used several properties for the expected value, namely E[U + V] = E[U] + E[V], E[aU] = a*E[U], E[b] = b, for random variables U,V and constants a,b. Note that Var(X) can also be written as E[X^2] - E[X]^2, which makes it easy to remember. I hope this helped! Please write back if you have further questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/