$X^TXw$ is normal?
I'm reading a paper in which it claims that if the matrix $X in mathbb{R}^{n times n}$ has elements which are normal and independent, then for an arbitrary vector $w$, $X^TXw$ is distributed as $sqrt{n}||w||N(0,I_n)$. Is it true? Because multiplying a normal to itself does not produce a normal. (http://papers.nips.cc/paper/6796-learning-relus-via-gradient-descent.pdf, page 7, equation 6.3)
probability-theory normal-distribution concentration-of-measure
asked Nov 27 at 5:25
S_Alex
858
add a comment |
I'm reading a paper in which it claims that if the matrix $X in mathbb{R}^{n times n}$ has elements which are normal and independent, then for an arbitrary vector $w$, $X^TXw$ is distributed as $sqrt{n}||w||N(0,I_n)$. Is it true? Because multiplying a normal to itself does not produce a normal. (http://papers.nips.cc/paper/6796-learning-relus-via-gradient-descent.pdf, page 7, equation 6.3)
probability-theory normal-distribution concentration-of-measure
asked Nov 27 at 5:25
S_Alex
858
This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23
add a comment |
I'm reading a paper in which it claims that if the matrix $X in mathbb{R}^{n times n}$ has elements which are normal and independent, then for an arbitrary vector $w$, $X^TXw$ is distributed as $sqrt{n}||w||N(0,I_n)$. Is it true? Because multiplying a normal to itself does not produce a normal. (http://papers.nips.cc/paper/6796-learning-relus-via-gradient-descent.pdf, page 7, equation 6.3)
probability-theory normal-distribution concentration-of-measure
asked Nov 27 at 5:25
S_Alex
858
I'm reading a paper in which it claims that if the matrix $X in mathbb{R}^{n times n}$ has elements which are normal and independent, then for an arbitrary vector $w$, $X^TXw$ is distributed as $sqrt{n}||w||N(0,I_n)$. Is it true? Because multiplying a normal to itself does not produce a normal. (http://papers.nips.cc/paper/6796-learning-relus-via-gradient-descent.pdf, page 7, equation 6.3)
probability-theory normal-distribution concentration-of-measure
probability-theory normal-distribution concentration-of-measure
asked Nov 27 at 5:25
S_Alex
858
asked Nov 27 at 5:25
S_Alex
858
edited Nov 27 at 6:33
asked Nov 27 at 5:25
S_Alex
858
asked Nov 27 at 5:25
S_Alex
858
asked Nov 27 at 5:25
S_Alex
858
858
This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23
add a comment |
This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23
This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23
add a comment |
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This doesn't sound right. When $w=(1,0,ldots,0)^T$, the first entry of $X^TXw$ is Chi-squared, not normal.
– user1551
Nov 27 at 5:42
papers.nips.cc/paper/…, Page 7, equation 6.3: $(w^*)^TX^TX$ is replaced with $a^T$, which is Gaussian.
– S_Alex
Nov 27 at 6:23