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seen May 21 at 12:15

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!