89 reputation
7
bio website
location Montreal, Canada
age 23
visits member for 2 years, 2 months
seen Dec 10 at 7:57

Math and physics student


Sep
24
awarded  Autobiographer
Aug
7
comment Splitting Integral into Two Parts
Thank you very much for your help and patience
Aug
7
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?
Aug
7
accepted Splitting Integral into Two Parts
Aug
7
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...
Aug
7
asked Splitting Integral into Two Parts
Jul
2
awarded  Curious
May
9
revised Delta function?
edited tags
May
9
revised Delta function?
added 13 characters in body
May
9
asked Delta function?
Jan
8
accepted Square of Bernoulli Random Variable
Jan
8
comment Square of Bernoulli Random Variable
Alright thanks a lot.
Jan
8
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?
Jan
8
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} $
Jan
8
revised Square of Bernoulli Random Variable
added 25 characters in body
Jan
8
asked Square of Bernoulli Random Variable
Mar
24
comment Bounded Linear Operators
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.
Mar
24
comment Bounded Linear Operators
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..
Mar
24
comment Bounded Linear Operators
Does this mean my steps are wrong?
Mar
24
accepted Bounded Linear Operators