meta user
network profile
| bio | website | neblink.wordpress.com |
|---|---|---|
| location | Montreal, Canada | |
| age | 29 | |
| visits | member for | 1 year, 9 months |
| seen | 8 hours ago | |
| stats | profile views | 533 |
I'm a graduate math student at Concordia University, Montreal.
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Jun 27 |
comment |
Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous @CameronBuie Thanks for your edit. You read my mind. I don't understand why, though, the $\arg_{b \in B}$ notation is not equivalent to the $B_a$ notation. |
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Jun 27 |
asked | Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous |
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Jun 23 |
revised |
Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$ change X_t, Y_t to X,Y, added explanation of quasilinear, per Davide comment and did answer |
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Jun 23 |
comment |
Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$ Strong as always. Thank you did. It's true that stochastic processes are not revelant. I got confuse in adapting my problem. I will correct the question. |
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Jun 23 |
accepted | Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$ |
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Jun 22 |
comment |
Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$ @DavideGiraudo I mean it's both quasiconcave and quasiconvex. Maybe only one of these is possible though. |
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Jun 22 |
asked | Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$ |
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Jun 19 |
revised |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure 1 char in body, following Byron comment, so that all $t$ are nonnegative |
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Jun 19 |
comment |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure @ByronSchmuland Good point. I forgot to mention that the $t$ are positive. Thank you. |
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Jun 19 |
comment |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure Both your answer and the one of copper.hat are excellent. I throw a coin to decide the winner. |
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Jun 19 |
accepted | Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure |
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Jun 19 |
comment |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure @copper.hat I meant the integral of $x^t$, sorry. |
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Jun 19 |
revised |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure 2 char in title and body, following Nimza comments |
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Jun 19 |
comment |
Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure @Nimza Good eyes, I should have wrote $x^t$, not $x$. Thank you. |
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Jun 19 |
asked | Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure |
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Jun 15 |
accepted | Finding the fixed point of a function |
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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? |
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Jun 14 |
comment |
Finding the fixed point of a function which theorem do you have in mind? |
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Jun 14 |
asked | Finding the fixed point of a function |
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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 $|g|\le 1$ |
