I IvanCarrie New member Joined Jan 2, 2019 Messages 4 Jan 8, 2019 #1 Given \(\displaystyle U=\frac{X-EX}{\sigma_X} \) so \(\displaystyle E = 0 \) how does \(\displaystyle E[U^2] = 1 \) ?

Given \(\displaystyle U=\frac{X-EX}{\sigma_X} \) so \(\displaystyle E = 0 \) how does \(\displaystyle E[U^2] = 1 \) ?

tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 10,675 Jan 8, 2019 #2 IvanCarrie said: Given \(\displaystyle U=\frac{X-EX}{\sigma_X} \) so \(\displaystyle E = 0 \) how does \(\displaystyle E[U^2] = 1 \) ? Click to expand... \(\displaystyle U^{2} = \left(\dfrac{X - E[X]}{\sigma_{X}}\right)^{2}\) --Can you prove it?

IvanCarrie said: Given \(\displaystyle U=\frac{X-EX}{\sigma_X} \) so \(\displaystyle E = 0 \) how does \(\displaystyle E[U^2] = 1 \) ? Click to expand... \(\displaystyle U^{2} = \left(\dfrac{X - E[X]}{\sigma_{X}}\right)^{2}\) --Can you prove it?

H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 6,029 Jan 8, 2019 #3 \(\displaystyle \frac{3}{3{\int_{-1}^1 x dx= 0\) but \(\displaystyle \frac{3}{2}\int_{-1}^1 x^2 dx= 1\).

\(\displaystyle \frac{3}{3{\int_{-1}^1 x dx= 0\) but \(\displaystyle \frac{3}{2}\int_{-1}^1 x^2 dx= 1\).