Given U=\frac{X-EX}{\sigma_X} so E[U] = 0 how does E[U^2] = 1 ?
I IvanCarrie New member Joined Jan 2, 2019 Messages 4 Jan 8, 2019 #1 Given U=X−EXσX\displaystyle U=\frac{X-EX}{\sigma_X} U=σXX−EX so \(\displaystyle E = 0 \) how does E[U2]=1\displaystyle E[U^2] = 1 E[U2]=1 ?
Given U=X−EXσX\displaystyle U=\frac{X-EX}{\sigma_X} U=σXX−EX so \(\displaystyle E = 0 \) how does E[U2]=1\displaystyle E[U^2] = 1 E[U2]=1 ?
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Jan 8, 2019 #2 IvanCarrie said: Given U=X−EXσX\displaystyle U=\frac{X-EX}{\sigma_X} U=σXX−EX so \(\displaystyle E = 0 \) how does E[U2]=1\displaystyle E[U^2] = 1 E[U2]=1 ? Click to expand... U2=(X−E[X]σX)2\displaystyle U^{2} = \left(\dfrac{X - E[X]}{\sigma_{X}}\right)^{2}U2=(σXX−E[X])2 --Can you prove it?
IvanCarrie said: Given U=X−EXσX\displaystyle U=\frac{X-EX}{\sigma_X} U=σXX−EX so \(\displaystyle E = 0 \) how does E[U2]=1\displaystyle E[U^2] = 1 E[U2]=1 ? Click to expand... U2=(X−E[X]σX)2\displaystyle U^{2} = \left(\dfrac{X - E[X]}{\sigma_{X}}\right)^{2}U2=(σXX−E[X])2 --Can you prove it?
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 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 \(\displaystyle \frac{3}{2}\int_{-1}^1 x^2 dx= 1\).