c*****t
发帖数: 520 | 1 Thank you very much for your reply. But I do not understand this conclusion.
I think it is too strong. I want to know it in detail. Can you recommend
some books on this topic?
In the paper "The Approach of Solutions of Nonlinear Diffusion Equations to
Travelling Front Solutions", by Paul C. Fife and J.B. McLeod in 1977, the
authors discussed the following equation:
u_t=u_xx+f(u), -\infty0.
f\in C^1[0,1], f(0)=f(1)=0.
If the initial data u(x,0)\in[0,1], then u(x,t)\in[0,1] for any t>... 阅读全帖 |
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c*******h
发帖数: 1096 | 2 K_1(0)=infty
never mind, i found a solution from somewhere. the answer is infty. |
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a***s
发帖数: 616 | 3 This seems impossible except some degenerated cases.
Let $Y_k := K(x^*, X_k)$. Since $\{X_k\}_{k \geq 1}$ are i.i.d., $\{Y_k\}_{k
\geq 1}$ are also i.i.d. Since the value of the kernal is between 0 and 1,
$Y_k$'s are nonnegative. Then Borel-Cantelli lemma says $\sum_{k=1}^{\infty}
Y_k = \infty$ with probability one unless $Y_k = 0$ with probability one.
Therefore for the infinite sequence $K$ to be in $l^1$ with positive probabi
lity (one, actually), we must have $Y_k = 0$ with probability one. ... 阅读全帖 |
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a***s
发帖数: 616 | 4 Let $\{ X_k \}_{k \geq 1}$ be i.i.d. random variables such that they are non
negative and $\Pr( X_1 > 0 ) > 0$.
In particular, we can assume $\Pr( X_1 > c ) = \delta > 0$ with some nonrand
om positive constants $c$ and $\delta$.
Now B-C lemma tells us that with probability one, events $\{ X_k > c \}$ hap
pens infinitely many times. Thus with probability one, $\sum_{k=1}^{\infty}
X_k = \infty$. |
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a***n
发帖数: 3633 | 5 如果知道f \in L_1
而且\int_0^\infty{sin(t)*f(t)dt}=0
\int_0^\infty{cos(t)*f(t)dt}=0
能不能说f=0? |
|
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f**********d
发帖数: 4960 | 7 设f(x)=\int_{-\infty}^{x} N(0,1) dt, 其中N(0,1)是标准正态分布，
则f(x)在从-infty到0上的积分有限么？ |
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M****o
发帖数: 4860 | 8 f是[0,\infty)上严格单调增，非负函数
是不是一定存在另一个[0,\infty)上严格单调增，非负函数g
使得
1) g locally Lipschitz (or even smooth?);
2) g(s)<=f(s) (or g(s)>=f(s)) for all s>=0;
3) f(0)=g(0)? |
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M****o
发帖数: 4860 | 9 f是[0,\infty)上严格单调增，非负函数
是不是一定存在另一个[0,\infty)上严格单调增，非负函数g
使得
1) g locally Lipschitz (or even smooth?);
2) g(s)<=f(s) (or g(s)>=f(s)) for all s>=0;
3) f(0)=g(0)? |
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w****1
发帖数: 4931 | 10 It looks like you need a lesson in calculus. The divergence of perturbation
series has nothing to do with renormalization.
I suggest you do the following exercise. Consider the integral \int_{-\infty
}^\infty dx \exp(-x^2 - g x^4).
Calculate it perturbatively in powers of g, and show that the series in
powers of g has zero radius of convergence. Next, show that the if you cut
off the series at the point where it starts to diverge, the error is
comparable to the contribution from instantons, whic |
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p*****k
发帖数: 318 | 11 for powers of matrix, you normally want to diagonalize the matrix first:
say P'JP = diag{j1,j2,....jm}, where P'=P^(-1), and j's are the eigenvalues of J.
if i understand your question correctly, you want to compute the limit of (I-J/n)^n when n->infty (where I is the identity matrix)?
insert bunch of PP'(=I) in the product:
(I-J/n)^n = PP'(I-J/n)PP'(I-J/n)...PP'(I-J/n)PP'=P{[P'(I-J/n)P]^n}P'
=P[diag{(1-j1/n)^n,(1-j2/n)^n,...,(1-jm/n)^n}]P'
so seems to me the limit when n->infty is:
P[diag{e^(-j |
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i****e
发帖数: 78 | 12 maybe the question is (1-J/n)^m as m-> infty? then positive real parts may
be related to the convergence and rate.
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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H****y
发帖数: 19 | 13 It's not (I-J/n)^n when n->infty, it should be Prod_n^infty (I-J/n)
If you take a ln, then the i^th diagonal element becomes Sum_n ln(1-Ji/n).
If you approximate ln(1-Ji/n) by -Ji/n, then it becomes Sum_n (-Ji/n), which
does not converge.
One can confirm this argument with Mathematica, using Ji=1, and calculate
Sum_n ln(1-1/n). You'll find the sum does not converge. |
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Q***5
发帖数: 994 | 14 Thanks.
From the hint, I guess we need construct a function f>0 on [0,1), such that,
\int_0^1 f < \infty, and \int_0^1 g = \infty, where
g(t) =(1/(1-t)) * \int_t^1 f(x)dx for any t in [0 1)
I was tempted to consider f = 1/(1-x)^a, where 0
, because the correspond g has finite integeral.
I guess we need something like: f = h(x)/(1-x), where h(x) goes to 0 as x
goes to 1, fast enough to make f finitely integerable, but slow enough so as
to make the corresponding g to |
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c*******d
发帖数: 255 | 15 assume 2*N flips, and we'll let N->infty
the probability of having N heads and N tails is
Pr = C(2N, N)*p^N*(1-p)^N
= (2N)!/(N!)^2 *[p(1-p)]^N
using sterling formula, we have
Pr = [sqrt(2 pi 2N)*(2N/e)^(2N)*(1+O(1/N))] /
[sqrt(2 pi N) * (N/e)^N*(1+O(1/N))]^2 * [p(1-p)]^N
~= 1/sqrt(pi N) * [4p(1-p)]^N
since 4p(1-p) <= 1, we have [4p(1-p)]^N <=1
thus Pr -> 0 as N-> infty |
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j*****4
发帖数: 292 | 16 order statistics. for standard normal,
assume X_{1,n}
E(X_{k,n})=\int_{-infty}^{+infty}xF(x)^(k-1)(1-F(x))^(n-k)f(x)dx
where f is pdf and F is cdf.
it's easy to derive result for normal dist. |
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p*****k
发帖数: 318 | 17 mathmean, judging from what you described, seems Crack was
talking about d(log C)/dT = - Theta / C,
which goes to 0 when S->+infty (deep ITM),
goes to +infty when S->0 (deep OTM)
for a non-dividend stock
[edited: the limit when S->0 is tricky] |
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O********9
发帖数: 59 | 18 第三题有个比较接单的解法:
注意到 N^2 a^N 是 Gamma 分布的核。由Gamma分布定义，对任意正实数k 和 b
int_0^{\infty} N^{k-1} e^{- N/b} dN = b^k Gamma(k)
因为
N^2 a^N = N^(3-1) e^{- N ln(1/a)}
所以，
int_0^{\infty} N^2 a^N dN = (1/ln(1/a))^3 Gamma(3) = 2 (1/ln(1/a))^3。
(这里 Gamma(k)=(k-1)!)
best
option,
became
got 30.
price? |
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c*********g
发帖数: 154 | 19 略微有一点错，应该是：
Pr[N>=n]=1/n*(n-1)!
E[N]=sum{from n=1 to infty} n*Pr[N>=n] = sum{from n=1 to infty} 1/(n-1)! = e |
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f********y
发帖数: 278 | 20 Gaussian integral 是 \int_-\infty^\infty e^(-t^2)dt=\Pi^0.5
这是从负无穷积分到正无穷，如果给定任意两个值 a 和 b,那个积分从 a 到 b是不是
没有解析解的？
Thanks! |
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f********y
发帖数: 278 | 21 P(X>Y)=\int_0^{\infty} f(x)*P(x
=\int_0^{\infty} f(x)*P(x
=... |
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d*e
发帖数: 843 | 22 (2) might not be enough, we might need \int_0^\infty c^2(s)ds<\infty?
Given variance goes to 0, and I(t) is normally distributed, how to show I(t)
->0 in almost surely (in prob. 1)?
I appeciated all the above answers... thanks
it |
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x******a
发帖数: 6336 | 23 from either the fact that exp(sB_t-s^2t/2) is a martingale
or direct calculation, we can get
Eexp(sB_t)=exp(s^2t/2)=\sum_0^\infty s^(2n)*(t/2)^n/n!.
On the other hand,
Eexp(sB_t)=sum_0^\infty (s^nEB_t^n)/n!.
Compare the coefficients of the two series,
EB_t^(2n)/(2n)! = (t/2)^n/n!,
EB_t^(2n)=(2n)!(t/2)^n / n!.
n=3 ==> EB_t^6=... |
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w**********y
发帖数: 1691 | 24 大意了..错了..假设你赢的概率是p,那么
p=1/6+5/6(1-p) 所以p=6/11
这个看multivariate transformation of random variable
或者直接积分啊..
P(x1+x2
. |
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w**********y
发帖数: 1691 | 25 说说俺的理解:
Stochastic Calculus的顺序是这样的:
1. 定义了Brownian Motion: W_t
2. 仿照黎曼积分的方式定义了基于BM的Ito 积分 (取每个区间的前端点): \int_0^t f
(w,t) dW_t
这时候问题就来了,到底什么样的f才能用来定义Ito积分..当然f必须是adapted
measurable function.然后我们希望定义这个积分的expectation和variance都存在的
话, 就定义出了一个空间 L2(dp*dt) = {f: E(\int_0^t f^2(w,t) dW_t)<\infty}
这个时候出现一个问题就是: 我们当然希望连续函数f都可以很好的定义ito积分.但是
举个例子,f=exp(W^4)就不在上面的L2空间里面..所以那个条件太强了..所以我们把这
个条件放宽,定义一个新空间,变成LOC2(dp*dt) = {f: P((\int_0^t f^2(w,t) dW_t)<
\infty)}=1.这个就定义出来了一个更general的空间..这个区别就是:新空间里面允许f可以有
很大很大的值... 阅读全帖 |
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L**********u
发帖数: 194 | 26 扯淡的人还真多。
这个问题就是直接积分，根本没有不需要用任何高深的东西。
under risk-neutral measure, the stock price satisfies
d(ln S)=rdt+\sigma dW,
therefore the pdf of y=ln S is
g(y)=1/\sqrt{2\pi \sigma^2 t}e^{\frac{-(y-\ln S_0-(r-\sigma^2/2)t)^2}{2\
sigma^2 t}}.
The price of the option is given by
e^{-rt}\int_{-\infty}^{+\infty}ye^yg(y)dy.
|
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M****i
发帖数: 58 | 27 I think that it is your interviewer should go home to go over the definition
of the stochastic integral. Remember that, at the very beginning, we define
the stochastic integral M_T=int_0^T V_t dW_t for the processes V which are
square integrable: E(int_0^T V^2 dt)<\infty. And then it can be shown that
the process (M_t)_{0<=t<=T} is a square integrable martingale. This is just
your case since E(int_0^T |W_t|^2 dt)=int_0^T E(W_t)^2 dt=T^2/2<\infty.
be |
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r**a
发帖数: 536 | 28 I can understand your statement for the perpetual options. For the perpetual
calls, the price is equal to the price of stocks (see Steve's method of
mathematical finance). But this is really different with the non-perpetual
call whose price is definitly not equal to the price of stocks. What I am
challeging the initial statement is for the non-perpetual american call it
is not correct. You can't ignore the max. Further, I do not think \tau\in[0,
T]\cup{\infty} is equivalent to saying \tau\in[0,\... 阅读全帖 |
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k*******n
发帖数: 180 | 29 对于如下的积分，有个简单的解法
I = int_0^\infty exp(-x^2 - a^2/x^2) dx
对a求导，然后用x=a/y做变量替换，得到
dI/da = Const×I ==> I=Const exp(a)
另a=0，得到Const其实就是Erf(infty)。结果就很简单了。
你这个问题只要变换一下就可以了。这个是教科书上的解法，不过我实在想不起来是哪本教科书了，等
想起来了再补上。 |
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k*****y
发帖数: 744 | 30 One should be careful when the substitution is not monotonic.
In your calculation, the two roots cover the range of u in (1, \infty) and
(1, 0) for v in (2, \infty), so the the integral should be the sum/
difference of the ones from the two substitutions.
scratching my head to find out what's I did wrong, wasting several papers,
still not clue, many thanks for the help. |
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r******g
发帖数: 13 | 31 Thank you very much, kinecty, zan
didn't realize the two roots in different range
then it should be the sum of the two roots substitution, it will cancel out
the second parts, get equation (2), why is the difference? in both roots,
the range for v is (2, \infty)? equation (7) gives that v is in (-\infty, -2), then u is negative
what's wrong with equation (3)? related to "the substitution is not
monotonic"? |
|
k*****y
发帖数: 744 | 32 Because the second one is from 1 to 0 when v is from 2 to \infty, if you swap them you will have a minus sign. You need to double check whether (2) is the difference.
Sorry. You also need to swap the 0 and \infty in (3). That is why we want to use u - 1/u instead of u + 1/u.
out
-2), then u is negative |
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k*******n
发帖数: 180 | 33 我实在是解不出来了，求详细步骤
就是
P_j = 0.5 + 1/pi \int_0^\infty Re[e^{-i\phi\ln K}f_j(\phi,x,v)/(i\phi)] d\
phi
这个是怎么得到的啊。。。我算了好几次，结果都是
P_j = 1/(2\pi) \int_0^\infty e^{-i\phi\ln K}f_j(phi,x,v)/(i\phi) d\phi
按照heston的说法，f_j(phi)也是nonlinearly depend on phi的，那前面那个1/2究竟
是怎么出来的？？？
我愿意paypal $10聊表心意。 |
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c**********e
发帖数: 2007 | 34
For the original question, var(X_i)=infty, but var(max(X_i,1))
So there is no need to worry about it.
Both infinity.
E\tau = ab. I do not remember the var. I'll check it late. |
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c**********e
发帖数: 2007 | 35
For the original question, var(X_i)=infty, but var(max(X_i,1))
So there is no need to worry about it.
Both infinity.
E\tau = ab. I do not remember the var. I'll check it late. |
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k*****y
发帖数: 744 | 36 Isn't it by definition?
E[ sin(Wt) ] = int_{-infty}^{infty} sin(x)*p(Wt = x) dx
where p(Wt = x) is even and sin(x) is odd. |
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n*******t
发帖数: 67 | 37 那一步不是 optional stopping theorem 么，为什么只需要 P(tau < infty) = 1 就
可以了？
另外，这个办法要说明 P(tau < infty) = 1 （这本身就不太显然，虽然不难证明），
然后用到 exponential martingale，然后取极限……有没有更 intuitive 的方法啊。
。。 |
|
x******a
发帖数: 6336 | 38 P(tau < infty) = 1 这个不够
让tau是一个brownian motion第一次达到1的时间，tau满足P(tau < infty) = 1，
但是Ew_\tau\ne0. 是不是？ |
|
n*******t
发帖数: 67 | 39 那一步不是 optional stopping theorem 么，为什么只需要 P(tau < infty) = 1 就
可以了？
另外，这个办法要说明 P(tau < infty) = 1 （这本身就不太显然，虽然不难证明），
然后用到 exponential martingale，然后取极限……有没有更 intuitive 的方法啊。
。。 |
|
x******a
发帖数: 6336 | 40 P(tau < infty) = 1 这个不够
让tau是一个brownian motion第一次达到1的时间，tau满足P(tau < infty) = 1，
但是Ew_\tau\ne0. 是不是？ |
|
f*********5
发帖数: 367 | 41 LS没看贴就回了，汗。。。
假定random walk的起始点是0，问最后ruin在-10的概率p.
Let a=(sqrt(5)-1)/2, one can show that M_t:=a^{S_t} is a martingale, where S
_t the current wealth of the gambler. Let tau be the stopping of either
reaching -10 or +infty.
Then 1=M_0=E[M_{tau}]=p*a^{-10}+(1-p)*a^{-infty}=p*a^{-10}.
Therefore p=a^10<1, since a<1.
如果问a是怎么找出来的话，就是先设一个未知量a使得M_t能是个martingale的，解方
程求出a的值。 |
|
B***y
发帖数: 83 | 42 still I am not sure about the meaning of "travelling wave" in this case.
Let's think of Xi's as a vector (X1, X2, ..., X_N), then usually for a travelling wave
we mean that : There is a solution of the differential equation X(t) = (X1(t),
X2(t), ... X_N(t)) such that as t approaches to -\infty (minus infinity), there
is a limit for X(t); and as t approaches to +\infty (plus infinity), X(t) also
approaches to another limit. That's why it is called travelling wave: it travels
between two ultimate |
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f*******d
发帖数: 339 | 43 I need to calculate at points r1, r2, ... r_max the function f(r)
given
by
f(r)= \int_0^\infty g(k) sinc(kr) dk, where sinc(x)=sin(x)/x, g(k)
is a tabulated function. Now I can certainly do the integral one by
one, but hoping to use FFT to improve efficience, I observed that this
can be rewritten as
f(r)=1/r \int_0^\infty g(k)/k dk, which is just the sine
transformation.
So, I used the program sinft given in numerical recipes to
do the calculation. However, it seems that the result is very
bad a |
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n******d
发帖数: 244 | 44 We need to solve $x^12=2^x$, first,
$x\neq 0$, when $x>0$, we rewrite it
as $f(y)=y^y=2^{-\frac{1}{12}}$, where
$y=\frac{1}{x}$, $y^y$ has only critical point
at $y=e^{-1}$, and we have $f(0+)=1$,
$f(+\infty)=\infty$.
Notice $0<2^{-\frac{1}{12}}<1$. So since
$f(e^{-1})<2^{-\frac{1}{12}}$,
we have 2 solution in this case.
Next suppose $x<0$, let $y=-\frac{1}{x}$, we
have $f(y)=y^y=2^{\frac{1}{12}}$, notice
$2^{\frac{1}{12}}>1$, we have
1 solution in this case.
Draw the graph of $f(y)=y^y$$, you |
|
D**u
发帖数: 204 | 45 Great. Here is a "probabilistic" solution assuming that we do not notice to
use f_j(t) := exp(-x_j*t) right away.
Assume Y_i is a Poisson arrival process with arrival rate x_i, then we know
that
E(Y_i) = 1/x_i, and E(min(Y_i,Y_j)) = 1/(x_i+x_j). Suppose the CDF of Y_i
and Y_j are P_i(t) and P_j(t), then easy to see the CDF of min(Y_i,Y_j) is
1-(1-P_i(t))*(1-P_j(t)). Then with a little computation, we see that
E(min(Y_i,Y_j)) = \int_0^{\infty} (1-P_i(t))*(1-P_j(t)) dt
= \int_0^{\infty} exp(-x_ |
|
J*******g
发帖数: 267 | 46 来自主题: Statistics版 - 一道概率题 P(A win)
=\sum_{k = 0}^{\infty} P(A win at time 2*k+1)
=\sum_{k = 0}^{\infty} (3/10)*(7/10)^{2*k}
=(3/10)*(1/(1-(7/10)^2))
=30/51
P(B win) = 1 - P(A win) = 21/51 |
|
y*****y
发帖数: 98 | 47 如果是[0,1] bounded outcome score, 最简单的直接transform到[-infty,infty]，然
后fit linear regression.
复杂点但更好的做法有, ordinal regression (McCullagh, 1980)； binomial-logit-
normal, coarsened data model (Lesaffre, 2007). |
|
z****e
发帖数: 702 | 48 这个积分的形式是这样的：
I=\int_R [N(x|0,1)\int_x^\infty N(y|0,1)dy]dx
描述起来就是：对一个标准正态分布y在(x,+\infty)上积分，积分下限x本身也是一个
标准正态分布。
这个积分看起来不是很难，但是却写不出解析式？牛人帮着看看？ |
|
f**********d
发帖数: 4960 | 49 【 以下文字转载自 Faculty 讨论区 】
发信人: freelikewind (像风一样自由), 信区: Faculty
标 题: 借地儿问个正态分布的问题。
发信站: BBS 未名空间站 (Wed Feb 13 20:41:27 2013, 美东)
设f(x)=\int_{-\infty}^{x} N(0,1) dt, 其中N(0,1)是标准正态分布，
则f(x)在从-infty到0上的积分有限么？
哪位知道？？？ |
|
k***p
发帖数: 115 | 50 Suppose 0
independent, both are binomial distributed.
Now (X-Y)/N goes to (2a-1)*p as N approaches infty.
let 0
as N approaches infty, the CDF of (X-Y)/N up to b is either 0 or 1,
depending on the relation between b and (2a-1)*p?
Thanks a lot!
|
|
