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<t>Let (m_1,m_2,...,m_k) be the column representation of the matrix and let s_j=\sum_{i=j}^k\ 1/m_{i,j}. My guess is that the matrix giving the extreme value for the sum on the left hand side is of the form m_j=(0,...,0,1/j,1/(j+1),...,1/k) (i think if it's true the induction wrt k might be used to ...

<t>sounds like one where playing with copulas might help if use smth like F(x,y)=xy+3\rho(1-x)(1-y)xy. or like thisLet x1 and x2 be independent N(0,1). Then we can think of u1=\Phi(x1) and u2=\Phi(x2) as of the given uniform variables. Now if we construct y1=x1 and y2=x2+\rho_1x_1 then z1=\Phi(y1) a...

<t>supernova87ap(s) is a payoff of the game when you roll twice with decision threshold s. So if you roll s or less first time you roll the die again. p(s) is a random variable so we have to calculate it's expectation which will give the fair price of the game (for s*). I_{A} is the indicator of eve...

Let d_1, d_2 be what we get in rolls 1 and 2 respectively. Let p(s)=I_{d_1<=s}d_2+I_{d_1>=s+1}d_1. We need to maximize Ep(s) wrt to s, which gives s^*=3 and Ep(s^*)=4.25.

Yurasounds like leningrad or is it of your own quant lyrics?

<t>The immigration office at my university told me that not so many companies had participated in this everify thing before the rule was released. And it's going to be real quick and easy for a company to join it. so if a bank pays to file a h1b application, it wouldn't a big problem to fill out a c...

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<t>I passed the quantExam as well, but the phone interview was full of technical questions of the same level of difficulty I had at the quantExam. Plus a human resource person told me I should expect technical questions. So I think different CS teams (or different people) follow different styles eve...

<t>Actually, there were fifteen rows with twelve people sitting on each. plus they put a few more tables (like four or five) at the back of the auditorium to accommodate the rest of the crowd. SoI would say 200 people took this test (out of which 30-40 were applying for the Summer Institute). I met ...

daveangelwhat I meant is you showed that a_n=log(n+1)/log(n) converges to 1 without stating the convergence rate, so your statement was not enough to claim that (a_n)^n goes to 1.((1+1/n)^n was a counterexample, as we know it converges to e.)exoticaIt looks right.

QuoteOriginally posted by: daveangeltherefore 1^infinity = 1.using the same argument (1+1/n)^n goes to 1.

QuoteOriginally posted by: 029 for the e^(infinity*ln1) term i would simply say zero times infinity is equal to zero.We still need to show that n(\log\log(n+1)-\log\log(n)) goes to zero.

<t>Another solution:We can pair all 9,000,000,000 numbers in the range in the following way: the first and the last, the second and the last but one, etc.So we get1+0+0+0+0+0+0+0+0+0 and 9+9+9+9+9+9+9+9+9+9 = 91; 1+0+0+0+0+0+0+0+0+1 and 9+9+9+9+9+9+9+9+9+8 = 91; etc.There are 4,500,000,000 pairs eac...

you may try www.econometriclinks.com