1,702 reputation
1820
bio website nicolasessisbreton.com
location Montreal, Canada
age 30
visits member for 3 years, 4 months
seen Dec 18 at 13:08

I'm a graduate math student at Concordia University, Montreal.


Jun
15
accepted Finding the fixed point of a function
Jun
14
comment Finding the fixed point of a function
I'm not clear on how to interpret mixed strategies in my case. The strategy sets $A$ and $B$ are both convex subset of $\mathbb{R}^n$. Let $\Pi(A)$ be the set of possible probability distribution on $A$. My understanding is that the mixed strategies are $\Pi(A) \times \Pi(B)$. Aren't these different?
Jun
14
comment Finding the fixed point of a function
which theorem do you have in mind?
Jun
14
asked Finding the fixed point of a function
Apr
12
accepted Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
Apr
11
awarded  Teacher
Apr
11
comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
@leo Good idea. It will enhance the post. I did it. Hope it's correct.
Apr
11
answered Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
Apr
10
comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
@DavideGiraudo I think any kind of inequality will do as long as it involves $\|f \chi_E \|_r$, $\|f \chi_{\tilde E}\|_s$ and $\|f\|_p$.
Apr
10
comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
@DavideGiraudo This point is not clear from the question. With the work I did, I think the inequality should be of the form $\|f \chi_E\|_r+ \|f\chi_{\tilde E}\|_s \le c \|f\|_p$ for some $c>0$.
Apr
10
revised Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
9 characters: added meaning of {\tilde E}
Apr
10
comment Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
@MattN. Sorry, it's $\mathbb R \setminus E$.
Apr
10
asked Find an inequality for $\|f\|_p$ when $f=f \chi_E+f\chi_{\tilde E}$ and $m(E)=m(\{x:|f|> 1\})<\infty$
Apr
4
comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
@lazyhaze I see, thanks. It suffices to take $g=f\chi_{\mathbb{R} \setminus E}$.
Apr
4
asked Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
Mar
9
comment Brownian motion: hitting time of double barrier vs hitting time of single barrier
Our textbook refers to these as the heat equations (Introduction to Stochastic Process, Lawler). The precision you add on this in your post, helps me better understand these equations. Thank you very much Didier.
Mar
9
comment Brownian motion: hitting time of double barrier vs hitting time of single barrier
how do you get $\partial_t u=\frac12 \partial_{zz}^2u$? I know the reflexion principle, and the Markov property of a Brownian motion, but neither seem to justify this statement.
Mar
8
comment Brownian motion: hitting time of double barrier vs hitting time of single barrier
For the continuity of $z\mapsto\mathrm P_z(T\gt1)$. This seems intuitive as a Brownian motion has continuous path, but without deriving the distribution of $P_z(T\gt1)$, can it be prooved?
Mar
7
revised Brownian motion: hitting time of double barrier vs hitting time of single barrier
change T_1 for T_x, as per original question, following Didier comment
Mar
6
revised Brownian motion: hitting time of double barrier vs hitting time of single barrier
corrected T_x to T_1 to match Didier answer (i think its clearer if both the question and the answer agrees)