A abc4616 New member Joined Sep 30, 2006 Messages 9 Oct 14, 2006 #1 If a random variable X is defined such that E[(X-1)^2]=10 and E[(X-2)^2]=6, find the mean and variance of X. Does anyone know how to approach this question??
If a random variable X is defined such that E[(X-1)^2]=10 and E[(X-2)^2]=6, find the mean and variance of X. Does anyone know how to approach this question??
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Oct 14, 2006 #2 Using the additive and homogeneous properties we get: \(\displaystyle \begin{array}{rcl} E\left[ {\left( {X - 1} \right)^2 } \right] = 10\quad & \Rightarrow & \quad E\left( {X^2 } \right) - 2E(X) + 1 = 10 \\ E\left[ {\left( {X - 2} \right)^2 } \right] = 6\quad & \Rightarrow & \quad E\left( {X^2 } \right) - 4E(X) + 4 = 6 \\ \end{array}.\) Solve the system for \(\displaystyle E\left( {X^2 } \right)\quad \& \quad E\left( X \right).\) Recall that \(\displaystyle V(X) = E\left( {X^2 } \right) - E^2 \left( X \right).\)
Using the additive and homogeneous properties we get: \(\displaystyle \begin{array}{rcl} E\left[ {\left( {X - 1} \right)^2 } \right] = 10\quad & \Rightarrow & \quad E\left( {X^2 } \right) - 2E(X) + 1 = 10 \\ E\left[ {\left( {X - 2} \right)^2 } \right] = 6\quad & \Rightarrow & \quad E\left( {X^2 } \right) - 4E(X) + 4 = 6 \\ \end{array}.\) Solve the system for \(\displaystyle E\left( {X^2 } \right)\quad \& \quad E\left( X \right).\) Recall that \(\displaystyle V(X) = E\left( {X^2 } \right) - E^2 \left( X \right).\)