user41725
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 Apr 9 comment $\rho_\gamma(X)=\frac{1}{\gamma} \log \mathbb{E}[e^{-\gamma X}]$ makes sense, thanks for your help and patience Apr 9 comment $\rho_\gamma(X)=\frac{1}{\gamma} \log \mathbb{E}[e^{-\gamma X}]$ would you please elaborate a bit on the fact that the inequality is strict for all non constant X. ( almost surely ). Thanks in advance! Apr 9 comment $\rho_\gamma(X)=\frac{1}{\gamma} \log \mathbb{E}[e^{-\gamma X}]$ Thanks! Am I assuming correctly that you used Jensen en.wikipedia.org/wiki/Jensen's_inequality $\frac{1}{\gamma}\log\mathbb{E}[e^{-\gamma 2 X}] \geq \frac{1}{\gamma}\log(\mathbb{E}[e^{-\gamma X}])^2$ Apr 9 comment $f(z):=\int_{\mathbb{R}} \frac{1}{t-z} d\mu(t)$ show $\lim_{y\rightarrow 0}iyf(iy)=-\mu(\lbrace 0 \rbrace)$ thanks so much! Apr 1 comment $f(z):=\int_{\mathbb{R}} \frac{1}{t-z} d\mu(t)$ show $\lim_{y\rightarrow 0}iyf(iy)=-\mu(\lbrace 0 \rbrace)$ Thanks again, but I have one more question. I don't understand the last equality, it is a bit too fast for me to see this. Could you perhaps explain that last step, please? Feb 13 comment zeta function and probability divisible by k and choosing squares but the "P" in the sum should be small p, right? Feb 13 comment zeta function and probability divisible by k and choosing squares thanks a lot :) Oct 7 comment $\sigma$-algebra generated by open sets coincides with $\sigma$-ring generated by open sets. oh my bad. all good :) Oct 7 comment $\sigma$-algebra generated by open sets coincides with $\sigma$-ring generated by open sets. thanks, that made it more clear. One follow up: what do you mean by $\sigma$-ring containing the "base set" is a $\sigma$- algebra. what is meant by base set. Oct 6 comment if X not finite then O is not a $\sigma$ - algebra thanks you very much Oct 6 comment if X not finite then O is not a $\sigma$ - algebra ah, in the case when both the set and it's complement are not finite, right? Oct 6 comment if X not finite then O is not a $\sigma$ - algebra I'd say one that is infinite in size Nov 13 comment linear equations - under modulo Thanks for this hint. I realized in different languages it's a different letter. I meant to say the solution set L or E or any name. Nov 13 comment linear equations - under modulo I changed my mistake. Thanks for notifying, and for your reply! Oct 31 comment $A+\lambda B$ is invertible Thanks for letting me know!