b******w
发帖数: 52 | 1
dollar.
t \in [ 0, \infty), no constraint on t! |
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w******w
发帖数: 92 | 2 Just use reflection principle. let k=2m-1.
1. P= (1/2)^(2m-1) * (2m-2)! / (m! (m-1)!)
E=\infty
2. P= p^m * (1-p)^((m-1) * (2m-2)! / (m! (m-1)!)
E=??? (might be very complicated) |
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r***w
发帖数: 35 | 3 No need to use maple, it is elementary, integrate the function f(x)=(1/2)^2x
^2 on [1,\infty], by L'Hospital's rule, you can get
\sum=[(ln2)^2+2*ln2-2]/[2*(ln2)^3]
Or, sum x^n=1/(1-x) (geometric) |
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c*******e
发帖数: 150 | 4 note that the process Y_t satisfies dY_t = T_t Y_t dW_t
it is called the stochastic exponential of (T_t)_{t\in[0,T]}, sometimes
denoted as \epsilon(T)_t
If (T_t)_{t\in[0,T]} satisfies the Novikov condition, i.e.
\mathbb{E}[exp(1/2 \int_0^T T_s^2 ds)] < +\infty
then Y_t is also called an exponential martingale, since it is a true
martingale on [0, T] |
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c**********e
发帖数: 2007 | 5 The final equation is simple.
16/pi int_0^infty 1/x exp(-x*x) dx
But it does not converge. The answer is positive infinity.
x2/
alpha |
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p*****k
发帖数: 318 | 6 as shown above, the first person has a winning prob of 1-N^N/(N+1)^N, i.e.,
his losing prob is N^N/(N+1)^N.
i guess you simulated N=6, which is (6/7)^6 and indeed agrees with your
numerical answer.
if N->infty, it approaches 1/e, which is approximately 0.368. |
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j**********7
发帖数: 4 | 7 Many thanks! It is quite clear and correct.
You are so kind :)
eigenvalues of J.
(I-J/n)^n when n->infty (where I is the identity matrix)?
J is diagonalizable http://en.wikipedia.org/wiki/Diagonalizable, the above procedure should work, no? |
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p*****k
发帖数: 318 | 8 thanks. the question makes sense now.
but you sure it does not converge? seems to me you are saying the log of
this infinite product is going to -\infty, so the product converges to zero,
no? |
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o*******r
发帖数: 131 | 9 问题转化为
非对称random walk，往左走一步概率为P=0.55，往右走一部Q=0.45
开始点为0，左边边界为 -10， 右边为正无穷大。
问 P(先到达-10) 的概率
然后直接套公式得
1 - (Q/P)^infty |
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f*******y
发帖数: 988 | 10 正解，然后解你不满意的话继续 6次方， 8次方 .....
道理上是和infty norm是一样的，无非max不能求导，理论性质很差
偶数次方什么的，问题就有很好的理论性质，保证可解而且是global的最优 |
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p*****k
发帖数: 318 | 11
should not it be: sqrt(2)*[the distance between the center and the starting position]?
if you don't believe apc999's clever argument, you could solve the trajectory, which is r(theta)=r0*exp(-theta). the distance traveled is then:
sqrt(2)* int{from 0 to infty} [dr/d(theta)] d(theta)
and you got the same answer. |
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p*****k
发帖数: 318 | 12 hmm, this is misleading. mathematically the expected "ratio" is a
well-defined question, but it does not really have a physical meaning.
let's say there are N (->infty) families, would you measure the ratio
of boy vs. girl for each family, then average? no. the sensible thing
to do is to get a weighted average of the ratio by the # of children
each family has. this is exactly what was asked in the original
problem: the averaged # of boys (and girls). |
|
|
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p*****k
发帖数: 318 | 15 not sure what exactly DuGu had in mind, but im guessing that
he wants the radii of these circles nonzero and bounded.
otherwise, e.g., the infinite str8 line R could be considered
as a circle with infinite radius:
(x-R)^2+y^2=R^2 with R->infty |
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D**u
发帖数: 204 | 16 Sorry that the link I gave might bring more confusions than it clears up for
you. The picture might be confusing in the sense that they draw it in black
and orange circles, but indeed they are
black circles (not sphere) and orange spheres (not circles, with all radii 0
Also the black circles are NOT drawn after picking some points on the orange
spheres. The black ones and oranges are drawn independently.
though
is
,
something |
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D**u
发帖数: 204 | 17 The R^3 analogue of Hopf fi bration is indeed surprising.
Here is my thoughts on the new question "how to partition R^3 into pairwise
non-parallel lines?".
We know that the hyperboloid of one sheet
(x^2/a^2 + y^2/b^2 - z^2/c^2 = 1)
can be partitioned with pairwise non-parallel lines.
If we partition R^3 into hyperboliods
x^2/a^2 + y^2/a^2 - z^2 = 1 (0
PLUS the z-axis,
and then partition each hyperboliod with line, then these lines should be
pairwise non-parallel. |
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p*****k
发帖数: 318 | 18
redtulips, that's a very nice solution.
the asymptotic result for large n is c*x-(1-c)+O(1/x^n).
the coefficient c is:
int{from 0 to infty} dt exp{-2*int{from 0 to t} ds [1-exp(-s)]/s}
~0.747598, which is indeed very close to 3/4.
the earliest reference is
A.Renyi, "On a One-Dimensional Problem Concerning Random Place Filling" (1958) |
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w**********y
发帖数: 1691 | 19 You are still confused...
First, definition of martingale is: E(X_t|F_s)=X_s, s
So you need to prove E[sin(W_t)-sin(W_s)|F_s]=0. Following your ito formula,
sin(W_t)-sin(W_s) = \int_{s}^{t}cos(W_u)dW_u - 1/2\int_{s}^{t}sin(W_u)du
The first term is ito integrals, whose expectation AND ALSO conditional
expectation is zero;
As for the second term, E(\int_{s}^{t}sin(W_u)du)=0, BUT!!! E(\int_{s}^{t}
sin(W_u)du|F_s) \neq 0 (should be s*sin(W_s)?? need check.).
If you don't understa... 阅读全帖 |
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p*****k
发帖数: 318 | 20
yes, that is correct. thx for pointing it out.
majia222, you want the # of flips to be even to keep the rear light
still on. note sum(n from 0 to infty) lambda^n/n! is exp(lambda),
so in order to get the sum of all the terms with even n, one could
add exp(-lambda) with the same even terms but negative odd terms.
thus the answer is hyperbolic cosine. |
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n****e
发帖数: 629 | 21 This is a nice solution and should be marked.
我前几天胡思乱想也考虑过这道题。事实上有几种情况：
1.如果是random walk(discrete), 那么给定boundary A,B 期望stopping time？
2.如果是biased，往上走概率p，那么给定boundary A,B 期望stopping time？
3.如果改成brownian motion(with/without drift) 那么给定boundary A,B 期望
stopping time?
如果把A->\infty, 就是面试很常见的题了。
都可以用martingale来解。先放在这里大家讨论一下，hehe
B) |
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w**********y
发帖数: 1691 | 22 zhucai mm给的ODE和Black Schole解法,给的是Perpetual American Put的结果,就是说
payoff=max(K-S_t,0), t from 0 to infty.的option的结果
参考:Paul Wilmott on Quantitative Finance Chapter 9-early exercise and
american options |
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k*******d
发帖数: 1340 | 23 不觉得simple啊，完整写出来得挺复杂的，求大牛们的简便做法
定义stopping time \tau = the first time B(t) = a+(a-b)t, 要求的就是P(\tau<\
infty).
Racall that M(t) = exp(\theta B(t) - 1/2 * \theta^2 t) is a martingale,
M(the min of \tau and t) is also a martingale, then
E[M(the min of \tau and t)] = E[M(0)] = 1; i.e.
E[exp(\theta B(min of \tau and t) - 1/2 \theta^2(min of \tau and t))] =
用indicator function 把它拆开
E[exp(\theta B(\tau) - 1/2 \theta^2 \tau) * Indicator function(\tau<=t)]
+ E[exp(\thata B(t) - 1/2 \theta^2 t) * Indicator |
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p*****k
发帖数: 318 | 24 this is just a shot in the dark:
would guess the question is from some d&c algorithm - one is
more interested in the asymptotic behavior of T when n->infty.
(T is probably only defined on N->N, so n/6 could be floor[n/6], etc)
then one needs the general version of the master theorem:
http://en.wikipedia.org/wiki/Akra-Bazzi_method |
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p*****k
发帖数: 318 | 25 i guess (s)he assumes "a" (b,c) is uniform on real by the limiting
procedure: a ~ U[-R,+R], then R->infty
on the other hand, if they are standard normals, there is a nice
geometrical approach for the general case |
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p*****k
发帖数: 318 | 26 E[N]=e, as the c.d.f. has a nice geometrical interpretation
(volume of the corner of a (n-1)-dim hypercube), so
Pr[N>=n]=1/(n-1)!
thus E[N]=sum{from n=1 to infty} 1/(n-1)!=e |
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c******r
发帖数: 300 | 27 刚实习回来，最近有点闲。
X1,..., Xn独立同分布expo distributed.
(1) min(X1, ... ,Xn) 的分布 （这个简单）
(2) max(X1, ..., Xn)的均值
(3) Var(max(X1, ..., Xn))以及当n->\infty, 这个variance的极限。 |
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k*******d
发帖数: 1340 | 28 3. what is the relation between wiener process and martingale
Wienner Proc is a Martingale.
4. (Position:Model Review at Morgan Stanley) The technical interview
consisted of the following:
for a<0
before b?
|a|/(|a|+b)
What is the expectation of the first hitting time of a or b?
\infty
What is B_t^2 (super or submartingale)
Submartingale
For X and Y two independent exponential variables with parameters a and b,
compute P(X>Y).
Giv |
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s*******u
发帖数: 35 | 29 P(X>Y)=\sum_{i=0}^{\infty}\sum_{j>i}P(X=j)P(Y=i)
It can be simplified because of exponential distribution.
b, |
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x******a
发帖数: 6336 | 30 int_0^\infty I_{B_t>0}I_{t<1} dB_t
I_{X} is the charactersitc function of set X.
B_t is a Brownian motion |
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p******5
发帖数: 138 | 31 what is the probability P that I have to wait more than x mins?
That means no buses show up during [0, x].
The probability of A bus showing up during [0,x] is PA = 1- e^{-1/3 x},
and
the probability of B bus showing up during [0,x] is PB = 1- e^{-1/6 x}.
Then P = 1 - (PA + PB - PA*PB).
The expected waiting time is \int^{\infty}_0 { P dx} |
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M****i
发帖数: 58 | 32 For this question, the simplest method that I can find is to use a
representation theorem of continuous local martingale and the theorem of
iterated logarithm of Brownian motion. An advantage of this method is that
c(t) can be allowed to be stochastic. More precisely,
Let c(t) be a stochastic process and we write
f(t)=\int_0^t c(s)^2 exp(2*\int_0^s c(r)dr) ds,
g(t)=exp(-\int_0^t c(r)dr).
Assume that
i)f(t)<\infty, a.s. for every t>=0;
ii)f(t) tends to infinity when t tends to infinity.
iii)g(t)*... 阅读全帖 |
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x******a
发帖数: 6336 | 33 11. a. law of the iterated logarithm.
b. weak law of large numbers.
c. P(|B_t/t|> x)=P(|B_1|>\sqrt{t}x)\to 0 for any x>0.
13 E(S_t)=E(S_0). lim(S_t)=0 (assuming \sigma is a nonzero constant) because
sigma*B_t-\sigam^2t/2 ->-\infty. |
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r*****r
发帖数: 630 | 34 Can anyone explain the answer to problem 1 a little more? thank you.
so the limit as the number of people goes to infty is 1/e? |
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w**********y
发帖数: 1691 | 35 多谢分享.大概做了做..欢迎补充和指正.
- sqrt(i)=?
e^{\pi/4 i} or - e^{\pi/4 i}
- You and me roll a dice,first one gets a six wins. You roll first. what
is the probability of you winning?
P(I win) = P(Y !win and I win) = 6/11
- A stair of n steps. Each time you step up 1 or 2 steps. How many
different ways are there to reach the top? what is the asymptotic limit?
Fibonacci sequence ..limF(n)/F(n-1)==x for n>2, solve x, and F(n) ~ x^{n-1}
- Moment generating function of standard model.
statistic book…
- Write a si... 阅读全帖 |
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t*******y
发帖数: 637 | 36 第二题应该是6/11吧
能讲讲这个吗? - X1 and X2 are independent random variable with pdf f and g.
what is what is the pdf of X=X1+X2
Jacobian matrix for X1+X2 and X1-X2..
多谢分享.大概做了做..欢迎补充和指正.
- sqrt(i)=?
e^{\pi/4 i} or - e^{\pi/4 i}
- You and me roll a dice,first one gets a six wins. You roll first. what
is the probability of you winning?
P(I win) = P(Y !win and I win) = 5/6*1/6
- A stair of n steps. Each time you step up 1 or 2 steps. How many
different ways are there to reach the top? what is the asymptotic... 阅读全帖 |
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e**********n
发帖数: 359 | 37 The answer depends on the sign of u if you use the residue theorem. However
thinking of real numbers only,
use the substitution
1/x -> \int_0^\infty e^{-xy} dy
the new 2D integral is easy to evaluate. |
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w**********y
发帖数: 1691 | 38 回到你这个ill-posed(?)题目啊:
如果按照你给的这个W_t, t \in [0,1] 和一个 B_t, t \in (1,\infty),然后要定义
correlation..那么这个correlation 只能定义成 corr (W_t1, B_t2), t分别在各自的
范围里.
所以你的correlation当然取决于你选的时间点.而不是什么随机过程的correlation..
俺吃饭去了.草草写了点看法,难免有错.
说实话,被timol说俺的例子错了之后,我还真是认认真真仔仔细细的想了一下才敢确保
自己没错的,然后才挖了个坑给你...
玩poker的时候,俺最喜欢也是最需要的就是9个人中有一个开始觉得一切显而易见,然后
开始冲动的时候..去年Xmas的时候,俺就是这么5分钟从200块翻成了1000块的..共勉共
勉.. |
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Q***5
发帖数: 994 | 39 L2(dp*dt) = {f: E(\int_0^t f^2(w,t) dW_t)<\infty}
I guess you meant to write dt instead of dW_t here.
f |
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j********t
发帖数: 97 | 40 unichaos的思路很赞。按照这个思路可以进一步找到更优解, 3000/e.
首先，显然尽量满载可以降低损失率，所以每次都满载1000。
假设有N个苹果要从A点运到B点，AB之间距离占总里程的比例为x，也就是AB长1000x
mile。那么运送次数是N/1000. 损失率是(1000x) * (N/1000) / N = x.
如果把1000 mile 分成n段运输，记损失率为x1, x2, ... xn. 问题相当于让总的损失
率最低，从而转化成求最优化问题。
Max{ 3000 * (1-x1)(1-x2)...(1-xn) }
subject to x1 + x2 + ... + xn = 1
n=2, x1=x2=1/2, 3000 * 1/4
n=3, x1=x2=x3=1/3, 3000 * 8/27
...
n -> \infty, x_i = 1/n, \lim 3000 * (1-1/n)^n = 3000/e |
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d*e
发帖数: 843 | 41 a stochastic process I(t)~N(0,v(t)), i.e. normal distributed with mean 0 and
variance v(t). If $v(t)->0$ as t->infty, do we have I(t) goes to 0 a.s.? |
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w**********y
发帖数: 1691 | 42 俺不做这行..懂皮毛..
首先..tail的corr的定义有很多种,一种常见方式是: P(x>u|y>u) as u goes to infty.
Gaussian copula 的corr是0,也就是说假设一个company极端事件的发生跟另外一个公
司的极端事件是独立的..
但是,现实生活中,极端事件总是一连串发生的.当你有N个公司时,这个joint loss
function,在independent assumption和dependent structure的情况下,tail的prob差
别可以几十上百倍了.. |
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s*******0
发帖数: 3461 | 43 突然 想到 如果相关系数为灵
要 copula 干啥啊 直接边缘分布 就行了
copula 就是 解决 相关系数不为零 边缘分布不同的 联合分布函数的简单的方法
infty. |
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w**********y
发帖数: 1691 | 44 Ok. I was not clear, and might already confused you.
The correlation you mentioned is defined as Cov(X,Y)/\sqrt{var(X)var(Y)}
I was talking about the tail/extreme dependence, defined as lim P(X>u|Y>u)
as u goes to infty.
For Gaussian copula, the corrlelation of X and Y is rho, which is in the
copula function; But the tail dependence is ALWAYS zero.
But when default/extreme event happens, there should be some kind of chain
reactions, and maybe high tail dependence. Gaussian Copula totally ignores... 阅读全帖 |
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a*******1
发帖数: 1554 | 45 E(XY)=1
cor(X,Y)=1
我的想法是求E(XY)的时候要用到X,Y的联合分布，而它只在X>0的时候有定义（因为Y只
有在这个时候有定义），所以积分区间是0开始，另外Y里面有条件概率，所以要乘2，
于是
E(XY)=2int_{0}^{\infty}x^2\exp{-x^2/2}dx=1.
同样，cor(X,Y)要用他们的联合分布，而这也只在X>0的时候有定义，而此时X=Y，所以
cor(X,Y)=1. |
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x******a
发帖数: 6336 | 46 I don't know.
Could you show me how to find
int_0^\infty e^(-c/x - x^2/2)dx,
for c>0?
or tell me the first few steps to your answer?
Thank you. |
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s*******j
发帖数: 9 | 47 e = E(\sum_i=1^\infty 1_{S_i<1}) + 1
and P(S_i<1) = 1/i! |
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l*****y
发帖数: 56 | 48 Thanks. I guess I misunderstood the question.
Btw, I think e = E(\sum_i=1^\infty 1_{S_i<1}). |
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r**a
发帖数: 536 | 49 Why do you allow \tau=\infty? For american calls，the condition \tao\in [0,
T] is enough, isn't?
Further, if you change the payoff function \Phi(S(T)):= max{S(T)-K, 0}, you
change the whole thing in my opinion. Because here the function \Phi is a
convex function of S(T). But if you define \Phi(S(T)):= S(T)-K, \Phi is a
line function of S(T). A convex function is related to submartingale while
the line function is related to the martingale. They will affect the time
you reach the maximum of final ... 阅读全帖 |
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r**a
发帖数: 536 | 50 Is it real? I mean in real world there really are some options which have
the proper ty that \tau=\infty?
payoff |
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