Expectation of standardised random variable

IvanCarrie

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Jan 2, 2019
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4
Given

U=XEXσX\displaystyle U=\frac{X-EX}{\sigma_X}

so

\(\displaystyle E = 0 \)

how does

E[U2]=1\displaystyle E[U^2] = 1 ?
 
Given

U=XEXσX\displaystyle U=\frac{X-EX}{\sigma_X}

so

\(\displaystyle E = 0 \)

how does

E[U2]=1\displaystyle E[U^2] = 1 ?


U2=(XE[X]σX)2\displaystyle U^{2} = \left(\dfrac{X - E[X]}{\sigma_{X}}\right)^{2} --Can you prove it?
 
\(\displaystyle \frac{3}{3{\int_{-1}^1 x dx= 0\) but \(\displaystyle \frac{3}{2}\int_{-1}^1 x^2 dx= 1\).
 
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