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comment L2 Matrix Norm Upper Bound in terms of Bounds of its Column
you do have the inequality $||A||_2\leq||A||_F$. Furthermore this inequality is tight if $A$ is rank 1 or less, so I'm not sure you will be able to find a better bound without further information.
Feb
10
revised Proving Holder's inequality for Schatten norms
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Feb
10
revised Proving Holder's inequality for Schatten norms
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Feb
10
revised Proving Holder's inequality for Schatten norms
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Feb
10
asked Proving Holder's inequality for Schatten norms
Feb
5
accepted Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
Feb
3
comment Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
@A.S. no restrictions on $X$ because we can always set the interval to be $[0,0]$ correct?
Feb
3
comment Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
@A.S. ah! well I stand corrected.
Feb
3
comment Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
@A.S. To be honest I couldn't follow all of Arash's integral manipulations (which is why I provided my own solution once I came up with it), but I feel that subgaussianity (or at least sub-exponential tails) must be required for $\int_{\bf R}\exp(|tx|)d\mu_X$ to have any hope of existing.
Feb
3
revised Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
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Feb
3
answered Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
Feb
3
revised Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
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Feb
2
comment Value of $\lim_{h\rightarrow 0}\frac{a^h-1}{h}$
use difference of powers to expand the numerator
Feb
2
comment How to integrate with a matrix in the measure?
well you really just need to figure out what is meant by the notation $d^{2n}M$, maybe it's defined in your book or your professor's notes or something.
Feb
2
revised Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
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Feb
2
comment How to integrate with a matrix in the measure?
I don't know enough about physics to answer this, but $\text{tr}(M^k)=\sum_i\lambda_i^k$, so if you treat $d^{2n}M$ as just $d\lambda_1...d\lambda_n$ or the equivalent for real and complex parts, you might be able to break the integral up into a product of integrals.
Feb
2
asked Extending the trace inner product to all matrix (real) inner products
Feb
2
revised Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
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Feb
2
revised Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$
[Edit removed during grace period]
Feb
2
asked Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$