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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
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.
Jun
23
accepted Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
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.
Jun
22
asked Show $f(a)=\operatorname{E} \min(aX,Y)$ is a quasilinear function of $a$
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
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.
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.
Jun
19
accepted Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
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.
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
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.
Jun
19
asked Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure
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.