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d1i1m1o1n [39]
3 years ago
11

Let X1, . . . ,Xn be an i.i.d. random sample from a N(0, 1) population. Define Y1 = 1 n n X i=1 Xi , Y2 = 1 n n X i=1 |Xi| . Cal

culate E(Y1) and E(Y2), and compare them.
Mathematics
1 answer:
nalin [4]3 years ago
8 0

For each 1\le i\le n, E[X_i]=0, so that

\displaystyle E[Y_1]=E\left[\frac1n\sum_{i=1}^nX_i\right]=\frac1n\sum_{i=1}^nE[X_i]=0

Meanwhile,

\displaystyle E[Y_2]=\frac1n\sum_{i=1}^nE[|X_i|]

Each of the X_i have PDF

f_{X_i}(x)=\dfrac1{\sqrt{2\pi}}e^{-x^2/2}

for x\in\Bbb R. From this we get

E[|X_i|]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty|x|e^{-x^2/2}\,\mathrm dx=\sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\,\mathrm dx=\sqrt{\frac2\pi}

\implies E[Y_2]=n\sqrt{\dfrac2\pi}

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\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

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