84 reputation
6
bio website
location Montreal, Canada
age 23
visits member for 1 year, 10 months
seen 2 hours ago

Math and physics student


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
Mar
24
revised Bounded Linear Operators
added 4 characters in body