Simon Sehayek
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 Sep24 awarded Autobiographer Aug7 comment Splitting Integral into Two Parts Thank you very much for your help and patience Aug7 comment Splitting Integral into Two Parts Haha, thanks for spelling it out, I should have just done that instead of plugging into Mathematica. The same should be true for a function with integrand $g(\max(x_1,x_2))f(|x_2−x_1|)$ since $x_1$ and $x_2$ are exchanged within both subregions, right? Aug7 accepted Splitting Integral into Two Parts Aug7 comment Splitting Integral into Two Parts I don't think there is a step discontinuity in my function. For example, $\exp(|x_2-x_1|)$ is completely continuous, so it should work right?... Maybe I calculated the integrals wrong, although Mathematica is saying the two integrals are not the same... Aug7 asked Splitting Integral into Two Parts Jul2 awarded Curious May9 revised Delta function? edited tags May9 revised Delta function? added 13 characters in body May9 asked Delta function? Jan8 accepted Square of Bernoulli Random Variable Jan8 comment Square of Bernoulli Random Variable Alright thanks a lot. Jan8 comment Square of Bernoulli Random Variable OK so I guess this isn't true if I shift the mean to be 0 so that $P_X(x)=\begin{cases} 1-p &\text{if } x=-p\\p & \text{if } x=1-p \end{cases}$. How can you handle it then? Jan8 comment Square of Bernoulli Random Variable Is this still the case if the distribution is not defined on the points {0,1} anymore? For example, if I shift the mean to be 0 so that $P_X(x)=\begin{cases} 1-p &\text{if } x=-p\\p & \text{if } x=1-p \end{cases}$ Jan8 revised Square of Bernoulli Random Variable added 25 characters in body Jan8 asked Square of Bernoulli Random Variable Mar24 comment How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$? Ya your definition seems a lot easier to use... however, I probably need to use the definition given in class. I'll readjust my proof accordingly. Mar24 comment How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$? Word for word from my book: $T$ bounded operator then $\\ \|T\|=\sup\{|\langle Tf,g\rangle|:\|f\| \le 1, \|g\| \le 1\}$. I don't understand why this doesn't hold in this case.. Mar24 comment How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$? Does this mean my steps are wrong? Mar24 accepted How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$?