Mixing
Mean of the partial sum of the normalized Gaussian random vector
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Let $X_1, X_2, dots, X_d$ be $d$ independent $N(0, 1)$ random variables, and let $Y=frac{1}{|X|}X$, where $X=(X_1, X_2, dots, X_d)$. Let the vector $Z in mathbf{R}^{k}$ be the projection of $Y$ onto its first $k$ coordinates, and let $L=|Z|^2$. Show that $mathbf{E}(L)=k/d$.
probability statistics normal-distribution means
asked Jan 27 at 14:06
BobSavageBobSavage
183
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$begingroup$
Let $X_1, X_2, dots, X_d$ be $d$ independent $N(0, 1)$ random variables, and let $Y=frac{1}{|X|}X$, where $X=(X_1, X_2, dots, X_d)$. Let the vector $Z in mathbf{R}^{k}$ be the projection of $Y$ onto its first $k$ coordinates, and let $L=|Z|^2$. Show that $mathbf{E}(L)=k/d$.
probability statistics normal-distribution means
asked Jan 27 at 14:06
BobSavageBobSavage
183
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, dots, X_d$ be $d$ independent $N(0, 1)$ random variables, and let $Y=frac{1}{|X|}X$, where $X=(X_1, X_2, dots, X_d)$. Let the vector $Z in mathbf{R}^{k}$ be the projection of $Y$ onto its first $k$ coordinates, and let $L=|Z|^2$. Show that $mathbf{E}(L)=k/d$.
probability statistics normal-distribution means
asked Jan 27 at 14:06
BobSavageBobSavage
183
$endgroup$
Let $X_1, X_2, dots, X_d$ be $d$ independent $N(0, 1)$ random variables, and let $Y=frac{1}{|X|}X$, where $X=(X_1, X_2, dots, X_d)$. Let the vector $Z in mathbf{R}^{k}$ be the projection of $Y$ onto its first $k$ coordinates, and let $L=|Z|^2$. Show that $mathbf{E}(L)=k/d$.
probability statistics normal-distribution means
probability statistics normal-distribution means
asked Jan 27 at 14:06
BobSavageBobSavage
183
asked Jan 27 at 14:06
BobSavageBobSavage
183
asked Jan 27 at 14:06
BobSavageBobSavage
183
asked Jan 27 at 14:06
BobSavageBobSavage
183
asked Jan 27 at 14:06
BobSavageBobSavage
183
183
add a comment |
add a comment |
1 Answer
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Note that $1=|Y|^2=sum_{i=1}^d Y_i^2$. By symmetry (there's a lot to go around) $EY_i^2$ is independent of $i$ and so $EY_i^2 = 1/d$. But $L=|Z|^2=sum_{i=1}^kY_i^2$, so $EL=k/d$.
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Note that $1=|Y|^2=sum_{i=1}^d Y_i^2$. By symmetry (there's a lot to go around) $EY_i^2$ is independent of $i$ and so $EY_i^2 = 1/d$. But $L=|Z|^2=sum_{i=1}^kY_i^2$, so $EL=k/d$.
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
$endgroup$
add a comment |
$begingroup$
Note that $1=|Y|^2=sum_{i=1}^d Y_i^2$. By symmetry (there's a lot to go around) $EY_i^2$ is independent of $i$ and so $EY_i^2 = 1/d$. But $L=|Z|^2=sum_{i=1}^kY_i^2$, so $EL=k/d$.
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
$endgroup$
add a comment |
$begingroup$
Note that $1=|Y|^2=sum_{i=1}^d Y_i^2$. By symmetry (there's a lot to go around) $EY_i^2$ is independent of $i$ and so $EY_i^2 = 1/d$. But $L=|Z|^2=sum_{i=1}^kY_i^2$, so $EL=k/d$.
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
$endgroup$
Note that $1=|Y|^2=sum_{i=1}^d Y_i^2$. By symmetry (there's a lot to go around) $EY_i^2$ is independent of $i$ and so $EY_i^2 = 1/d$. But $L=|Z|^2=sum_{i=1}^kY_i^2$, so $EL=k/d$.
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
answered Jan 27 at 14:16
kimchi loverkimchi lover
11.5k31229
11.5k31229
add a comment |
add a comment |
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