Question

M(t)=e^(2t+32t^2) find the mean and variance

Answer 1

M(t)=E[e^tX], X is a random variable--what distribution does X have?

Answer 2

mean = \[M \prime(0)\]
\[M \prime(t) = (64t + 2) e ^{2t + 32t ^{2}}\]
\[M \prime(0) = 2 : the mean\]

Answer 3

\[M \prime \prime(t) = (64t + 2)^{2} e ^{2t + 32t ^{2}} + 64e ^{2t + 32t ^{2}}\]
\[M \prime \prime(0) = 4 + 64 = 68\]
\[\sigma ^{2} = M \prime \prime(0) - (M \prime(0))^{2}\]
variance =68-4
=64

Answer 4

This is a standard normal distribution and therefore you know immediately that mean=2 and variance = 32(2), having said that - good job

Answer 5

yes - you're right! clearly seen in coefficients of exponent. It's been a while since stats. I really wondered if I remembered this stuff. (I sort of impressed myself). Thanks for the opportunity to exercise.

Answer 6

Second actuarial exam over a _____ ago. :) Despite what they say, you really don't use it. Am I too old to be on this site?