m******e
发帖数: 89 | 1 不同segment 的 model score/probabilty 能较吗？
company 的risk model 有10 个segments. 我在做strategy 时，能把它们的score/
probabilty 合在一起做，还是要分开在每个segment 上做。
我觉得应该分开做，然后在合到一起。因为PROBABILITY 只在它们所代表的ＰＯＰＵＬ
ＡＴＩＯＮ 上有效。但看到过去的一些ＳＣＯＲＥ CUTOFF 是合在一起做的。
谢谢！ |
|
m******e
发帖数: 89 | 2 谢谢回复！ 这里 “alignment之后同样的Score值在每个Segment上对应相同的风险”
的风险是指 BAD RATE OR ODDS 吗？如果是指ODDS 或者 BAD RATE 的话，那么 if i
have original log odds or probabilty from logistic regression, can i
directly compare them? Thanks ! |
|
l*****a
发帖数: 559 | 3 是不是这么一个simulation。如果是的话，单纯用统计能否得出同样的答案。
void probabilty(int n,int x,unsigned long long& cnt, unsigned long long&
totalCnt){
if(n == 0){
totalCnt++;
if(x == 0){
cnt++;
}
return;
}
probabilty(n - 1, x - 1, cnt, totalCnt);
probabilty(n - 1, x + 1, cnt, totalCnt);
}
cout<<(double)cnt / (double)totalCnt< |
|
l*****a
发帖数: 559 | 4 是不是这么一个simulation。如果是的话，单纯用统计能否得出同样的答案。
void probabilty(int n,int x,unsigned long long& cnt, unsigned long long&
totalCnt){
if(n == 0){
totalCnt++;
if(x == 0){
cnt++;
}
return;
}
probabilty(n - 1, x - 1, cnt, totalCnt);
probabilty(n - 1, x + 1, cnt, totalCnt);
}
cout<<(double)cnt / (double)totalCnt< |
|
p*g
发帖数: 141 | 5 1.如果往左/右的概率是 p/1-p
这个程序应该怎么写?
2. 总觉得这种recursive的不如递推的好
3. 觉得那两行
probabilty(n - 1, x - 1, cnt, totalCnt);
probabilty(n - 1, x + 1, cnt, totalCnt);
不需要乘 0.5么
多谢 |
|
p*g
发帖数: 141 | 6 1.如果往左/右的概率是 p/1-p
这个程序应该怎么写?
2. 总觉得这种recursive的不如递推的好
3. 觉得那两行
probabilty(n - 1, x - 1, cnt, totalCnt);
probabilty(n - 1, x + 1, cnt, totalCnt);
不需要乘 0.5么
多谢 |
|
b********e
发帖数: 41 | 7 You cannot simply replace 1/2 with 1/8, bacause you can have 3 girls, 2g&1b,
1g&2b and 3boys, with probabilty 1/8, 3/8, 3/8 an 1/8.
Any couple have a 1/4 chance to have no children, that's 1/4.
And they have 3/4 chance to have 3 children, the chance that all of them are
girls is 3/4*1/8=3/32.
So the probabilty should be at least 1/4+3/32=11/32 which is > 8/29.
I got a numerical answer, 0.77, which is pretty large, but I think is
possible. |
|
a****e
发帖数: 150 | 8 那个真不能叫probabilty theory. probabilty theory还是要建立在measure theory
上的,讲的一般是以下内容:The strong laws of large numbers. Conditional
expectations. Discrete parameter martingales: convergence, stopping times,
and optional sampling theorems. Uniform integrability. Weak convergence:
characteristic functions and the central limit theorem. Elements of large
deviations. The ergodic theorem. 你列的课跟我们系master level的theory课是
一个难度的. |
|
发帖数: 1 | 9 a: hiding in 100th cave,
b: hiding in xth cave
c: 98 random failed search from 1th to 99th.
given that a is true, the probabilty of c is 1. given b is true, the
probability of c is 1/99.
so given c has already happened, the conditional probality of a is 0.01/(0.
01 0.01/99)=0.99.
銆鍦iysk() 鐨勫ぇ浣滀腑鎻愬埌: 銆br />
: 婵炲备鍓濆﹢浣衡偓娑崇畳缁诲啫
顫楅崒婊冭姵濞戞柨鐗炵槐锟闁绘粍婢樺﹢顏堝礄閾忕懓绠涘ù婊冾儎濞嗐垺娼婚悙
鐢垫暰闂br />
: 鍛箲濠р偓濞戞挸绉甸弬渚炊閻愬弶韬闁告帪鎷9濞戯拷
: 椤忓懐顦柟璁规嫹98濞戞搩浜濈粈濠囨晬鐏炲墽姊鹃柡鍫濐槹
: 婢规﹢宕氶懜锟解偓閸屾艾寤哄ù婊冾儎濞嗩晧, 闁革讣鎷00闁告瑩鏀辩粈濠
囨煂瀹嫭妗ㄩ柛娑滄閹洘绂... 阅读全帖 |
|
B*******1
发帖数: 2454 | 10 careercup上的，看得不是很懂。
You have a digital clock having 24 hr format of HH:MM:SS. You have to find
following probabilities:
1- Occurence of a particular value(let say 05) in SS
2- As 1st point we have to find probabilities for particular value in hour
and minute fields.
3- probalities for other combinations(particular HH:SS or MM:SS values)
You have to answer the probability in terms of digits occurence...for
example in the second's field.. at unit place 10 digits(0-9) are possible,
and in the tens ... 阅读全帖 |
|
n***y
发帖数: 2730 | 11 13: last night I did not have time to go to detail. Here are some quick
thoughts on this problem for after work entertaining.
1-----
This is very close to negative binormial distribution (see http://en.wikipedia.org/wiki/Negative_binomial_distribution). According to wiki, let r be number of failure (use wiki term, which means accept invitation here), then the mode for nbd should be at floor(pr/(1-p)). (p is probability of sucess, same as in wiki)
Now noticing
p(NB(r, p) = k) = p (r-1 failures in... 阅读全帖 |
|
n***y
发帖数: 2730 | 12 13: last night I did not have time to go to detail. Here are some quick
thoughts on this problem for after work entertaining.
1-----
This is very close to negative binormial distribution (see http://en.wikipedia.org/wiki/Negative_binomial_distribution). According to wiki, let r be number of failure (use wiki term, which means accept invitation here), then the mode for nbd should be at floor(pr/(1-p)). (p is probability of sucess, same as in wiki)
Now noticing
p(NB(r, p) = k) = p (r-1 failures in... 阅读全帖 |
|
b**********5
发帖数: 7881 | 13 在说说最后一道题吧
给一个dictionary， 里面a list of words， 比如 “abcd“， ”abe", "bbc", "
bdcea"
从这个dictionary里， 给一个random word
怎么random么？
从这个字典里， 你可以看到：
第一个letter， 是“a”， 有多少probability， 是“b“， 有多少probability
then if u pick "a", "b" follow "a" with certain probabilty. if u pick "b",
"b" follow "b" with certain probability...
所以implement getRandomWord时， 先generate a random number between 0 and 1，
然后看第一个letter，“a“ 为first letter的probability是30%， 如果generate
为 0-0.3的时候， 就pick ”a”。 如果“b”的probability为60%， 那么generate
为0.3... 阅读全帖 |
|
G****8
发帖数: 475 | 14 正解！我的数字与你的一些不符合，但不知为啥结果一样。
gusu 赢 a game 的概率：
(1) with 4 points: 0.9^4=0.6561
(2) with 5 points: (4 choose 1)*0.9^4*0.1=0.26244
(3) with 6 points: (5 choose 2)*0.9^4*0.1^2=0.06561
(4) with 8+ points(after duece): 0.014402195
8=8+2*k(k=0) points: (6 choose 3)*0.9^3*0.1^3*(2^k*0.9^k*0.1^k)*0.9^2
10=8+2*k(k=1) points: (6 choose 3)*0.9^3*0.1^3*(2^k*0.9^k*0.1^k)*0.9^2
..... etc,
So the probability to win with 8+ points is a 等比数列, sum of (6 choose
3)*0.9^3*0.1^3*(2^k*0.9^k*0.1^k)*0... 阅读全帖 |
|
p***y
发帖数: 32 | 15 times in the past.
This question is the base of modern Philosophy of Science. Karl Popper is the
first who gave a rational answer to this questions which did influenced a
generation of scientists. His answer is in his famous book "Conjecturesand
Refutations".
The answer using probabilty is well studied by the Vienna School of philopophy
in the 1930's, which contains the most brilliant thinkers and mathematicians
at the time, but they failed to provide a convincing rational solution.
About 机械唯心主义 |
|
w********d
发帖数: 275 | 16 mainly some thing as
probabilty model,
stocastics,
and optimization |
|
l******2
发帖数: 5522 | 17 Why not???!!!
probabilty theory from Statistical thermodynamics |
|
l******2
发帖数: 5522 | 18 Why not???!!!
probabilty theory from Statistical thermodynamics |
|
l*******s
发帖数: 36 | 19 小弟申请08fall EE的MASTER，在纽约附近，专业排名60+，九月份入学。
现在正在选方向，目前考虑了两个请大家帮忙出出主意。
1 telecommunication network。主要课程有 Local and Metropolitan Area
Networks, High—Speed Networks, Communication Networks, High performance
Switches and Routers, Internet Architecture and Protocols,Probabilty，
Wireless Information Systems Lab 等等
另一个是Computer Electronic Devices。主要课程有：Advanced Computer
Hardware Design, Introduction to VLSI Design, VLSI system, Analog and High
Frequency Amplifier Design, Introduction to RF |
|
n******t
发帖数: 4406 | 20 No. It is a probability of both events happens.
Let M_n = min{S_k: 1<= k <= n}, then
P(M_n <= -b) = P(M_n <= -b; S_n > -b} +P {S_n <= -b}
But the first probabilty is just P(S_n < -b}
To be rigorous, this is based on a discrete version of
reflection principle, with stopping time T:=min{n>=0: S_n <= -b}. |
|
jl
发帖数: 398 | 21 A random varible converge to -\infty in provability
是什么定义? 多谢!
A random varible converge to -\infty with probability 1 s
是什么定义? 多谢! |
|
p***c
发帖数: 2403 | 22 只有一个r.v.，收敛个屁啊
如果是一列
1. for all K>0, \lim_n P(X_n<-K)=1
2. P({w|\lim_n X_n(w)=-\infty})=1 |
|
|
B*********h
发帖数: 800 | 24 ☆─────────────────────────────────────☆
alla (alla) 于 (Sat Jul 15 12:49:12 2006) 提到:
在这个版上看了半天，还是不太明白financial engineering和quant的主要工作是不是
program trading.
我自己瞎琢磨了一些批program trading的stratedgy，主要是根据一些简单的combined
probabilty ,conditional probablity和probablity distribution来提高自己的胜率，
再加上meony management 基本上就是把股市当成个大赌场。实际操作的效果也不错。
但是我所知有限，很快就黔驴技穷了。我的感觉是我的stratedgy还有很多可以提高的地方，
但是不知道从哪里入手。我试着看一些paper，但是都是偏微分方程，概率积分，我完全
看不懂。连作者的目的都看不太明白。很多是研究pricing的，我觉得对我的用处不大。
想到学校里选一些这方面的课，或者索性花一年时间学个financial engin |
|
m*******s
发帖数: 758 | 25 T_k = first time to see consective k same values of one dice
what is the expection ?
It was discussed about consective k same value of "1" : T_k(1)
E[T_k(1)] = 1/6 (E[T_{k-1}(1)]+1) + 5/6(E[T_{k-1}(1)]+1+E[T_k(1)])
How about this question ?
hee T_k = min {T_k(1),T_k(2),...,T_k(6)}. |
|
J****x
发帖数: 37 | 26 My two cents.
Let x = T_k(1). Then,
x = 1/6^k * k + sum_{i = 1}^k 5 * (x + i) / 6^k
Denote by y the expected number we want to have, then
y = x / 6 |
|
m*******s
发帖数: 758 | 27
this result is trivial.
" y = x / 6
seems ok ? but reason ??? could you please the reason behind it ?
Does T_k(1), T_k(2), ... T_k(6) independent ?
What is the joint distribution of them ?
... seems not easy |
|
p*****w
发帖数: 82 | 28 stock在risk neutral probability measure下的growth rate是risk free rate。
volatility 是mean reverting process.， 他的risk neutral probabilty measure底
下你的growth rate 你怎么value 啊？ |
|
p**e
发帖数: 41 | 29 1, Assume you have 4 sets of porkers, which are the same, each containg 52
cards. Put them (i.e. 208 cards) into a line. What is the number of
possible permutations?
2, Random walk,starting from point 0, with probabilty 0.5 for +2 steps and
with probablity 0.5 for -1 steps, Q: what is the prob of hitting the point
-1.
3, In linear regression estimation, how to handle with those wild outliers ? |
|
r*****d
发帖数: 44 | 30 if p!=1/2, it should be 0 since non-typical sequences have vanishing
probabilty of occurance asymptotically. |
|
m******n
发帖数: 354 | 31 Okay, let's talk about the very first thing:
The strong law of large numbers does not work for capital market at all!
Consider a forward: with the same underlying stock price process, the arbitrage(-free) price is not the expectation price(under real probabilty)!
But the arbitrage(-free) price should be the enforced market price! So we
want to change it to some expectation, so that we can calculate it using
stochastic tools generally.
To this end, we do change of measure (here: the probability m |
|
s***e
发帖数: 267 | 32 Thanks, very helpful. I think I got it.
arbitrage(-free) price is not the expectation price(under real probabilty)!
measure,
, |
|
S*******s
发帖数: 13043 | 33 what is the probabilty for a Brownian motion with drift
dB(t)=udt+dW(t), B(0)=0,
hitting A before it reaching B, where A>0>B? |
|
D**u
发帖数: 204 | 34 This is more like a brainteaser than a research math problem.
Even for 面试题 which this board might favor, it is not
limited to classic probabilty or BS. |
|
y****d
发帖数: 432 | 35 ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
前面说明：
需要的童鞋请到我的签名档的博客查找！谢谢！发E-mail太累了！
觉得有价值的话可以顶一下，以便更多的人看到！谢谢！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
本合集包含内容（总有你想要的吧？！呵呵）：
A Basic Course in Probability Theory
A Course In Probability Theory.djvu
A First Course in Statistics for Signal Analysis
A Handbook of Statistical Analyses Using R
A Handbook of Statistical Analyses using SPSS
An Introduction to Probability and Random Processes
An Introduction to Probability and Statistical Inference
An Introduction to Probability Th... 阅读全帖 |
|
u**x
发帖数: 45 | 36 Suppose you meant
Sa={[na]| a b.t. R+, any n b.t. N}
Sb={[nb]| b b.t. R+, any n b.t. N} where [] is gauss func.
Sa+Sb=N (1) && Sa*Sb=0 (2)
iff
(a) A, B are irrational
and (b) 1/A+1/B=1.
Prove:
(1)&&(2)=>(b)
for a K b.t N, for any k<=K b.t. N, the Probabilty
of k being in Sa, P(k,Sa)=1/A when K-> pos. inf.;
P(k,Sb)=1/B when K-> pos. inf.;
P(k, Sa+Sb)=P(k,Sa)+P(k,Sb)-P(k, Sa*Sb)=1/A+1/B when K->pos. inf;
However P(k, N)=1 for any K. Thus 1/A+1/B=1.
=> (a)
suppose A=m/n, B=m/(m-n), where m,n b.t. N |
|
S*********g
发帖数: 5298 | 37 And, this problem is not a problem of TIME EVOLUTION.
It is a problem of the distribution of the probabilty in the real space. |
|
y****d
发帖数: 432 | 38 ★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
前面说明：
需要的童鞋请到我的签名档的博客查找！谢谢！发E-mail太累了！
觉得有价值的话可以顶一下，以便更多的人看到！谢谢！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
本合集包含内容（总有你想要的吧？！呵呵）：
A Basic Course in Probability Theory
A Course In Probability Theory.djvu
A First Course in Statistics for Signal Analysis
A Handbook of Statistical Analyses Using R
A Handbook of Statistical Analyses using SPSS
An Introduction to Probability and Random Processes
An Introduction to Probability and Statistical Inference
An Introduction to Probability Th... 阅读全帖 |
|
c********h
发帖数: 330 | 39 严谨的说，如果每个segment的distribution不一样，是不可以pool together的。
就算这些segment的observed variables are comparable，也不能避免有很多其他的
unobserved variables。
直接pool together会有很大的hidden bias |
|
f*******n
发帖数: 2665 | 40 可以合在一起做，但你先要做alignment, 也就是对每个segment 上的score作线性变化
，alignment之后同样的Score值在每个Segment上对应相同的风险。这样10个Segment就
可以在一起做Cutoff和Strtegy了。 |
|
|
m******e
发帖数: 89 | 42 我知道怎么做scaling 但是不知道怎么做aligment cross segments. 能给点详细的指
点吗？ 或者有什么书有相关的内容？
十分感谢！ |
|
|
f*******n
发帖数: 2665 | 44 风险是指 BAD RATE OR ODDS 吗？是。这里的值不是模型的预测值，而是实际值，所以
你第二个问题的答案是NO。 |
|
g**********l
发帖数: 214 | 45 还不是太清楚，望能解释一下。
suppose i have 3 segments, high risk, med risk, low risk, with bad rate 10%,
5%, 1%. each segment has its own model and scores.
请问您说的是用这3个bad rate (10/5/1%) 去 adjust/align these 3 sets of scores?
how do you do that?
thanks! |
|
f*******n
发帖数: 2665 | 46 举个例子，seg 1中的500分对应10%， 而seg 2中的300分对应10%，相同的风险，却有
不同的score 。如果你做了alignment,你得到的应该是在两个seg 里都是400分（make
up)10%， 这样你才能把两个seg放在一起做cutoff.
act bad% score (seg1) score (seg2)
10% 500 300
5% 100 40
1% 10 20 |
|
f*******n
发帖数: 2665 | 47 举个例子，seg 1中的500分对应10%， 而seg 2中的300分对应10%，相同的风险，却有
不同的score 。如果你做了alignment,你得到的应该是在两个seg 里都是400分（make
up)10%， 这样你才能把两个seg放在一起做cutoff.
act bad% score (seg1) score (seg2)
10% 500 300
5% 100 40
1% 10 20 |
|
m******e
发帖数: 89 | 48 我想我大概明白了。问题是如果我在１０% 上calibrate two segments, how can i
make sure they are also aligned at 5%, 1% ?
Thans a lot!
make |
|
g**********l
发帖数: 214 | 49 请问具体是怎么做alignment呢？
用那个你们说的做例子
actual 10% bad rate is group/segment-wide metric.
500分 is individual-wide metric.
这两个怎么凑在一起呢
还有，另外两个segments have only 5% and 1% bad rate (as a segment-wide
metric). they can never have 10% bad rate as a grTFFFoup. sorry i am not
seeing how to align them at 10%.
thanks a lot！ |
|
