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 22h 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 deleted 1 character in body Feb 10 revised Proving Holder's inequality for Schatten norms deleted 1 character in body Feb 10 revised Proving Holder's inequality for Schatten norms added 24 characters in body 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 ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$ added 2 characters in body 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 ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$ deleted 18 characters in body 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 ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$ added 16 characters in body 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 ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$ deleted 1 character in body 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 ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$