o****e
发帖数: 80 | 1 pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
接。
我一道道加把
1。bionomial distribution近似
http://www.mitbbs.com/article/Quant/31175297_0.html
这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
np, then (n,k)*p^k*(1-p)^(n-
k)--->lamda^k*exp(-lamda)/k!
可是这样还是心算不出来阿？ 大虾们说说把
2. random walk
http://www.mitbbs.com/article_t/Quant/19184699.html | s***0
发帖数: 2 | | r*******y
发帖数: 290 | 3 1. use central limit theorem, the sum of 100 i.i.d. r.v. are close to normal
let x=1 if head and -1 if tail, p=0.5
then sum(xi) ~ N(0, 100), P(heads < 40) = P(sum of xi < -20) which is the
left tail of 2*sigma, if you still remembers the normal districution
the 2*sigma is about 2%, that's the answer
【在 o****e 的大作中提到】
: pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
: 接。
: 我一道道加把
: 1。bionomial distribution近似
: http://www.mitbbs.com/article/Quant/31175297_0.html
: 这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
: np, then (n,k)*p^k*(1-p)^(n-
: k)--->lamda^k*exp(-lamda)/k!
: 可是这样还是心算不出来阿？ 大虾们说说把
: 2. random walk
| o****e
发帖数: 80 | 4 thank you, who can explain sth about prob 3 and 4?
normal
【在 r*******y 的大作中提到】
: 1. use central limit theorem, the sum of 100 i.i.d. r.v. are close to normal
: let x=1 if head and -1 if tail, p=0.5
: then sum(xi) ~ N(0, 100), P(heads < 40) = P(sum of xi < -20) which is the
: left tail of 2*sigma, if you still remembers the normal districution
: the 2*sigma is about 2%, that's the answer
| n****e
发帖数: 629 | 5 最后一题
用Markov chain
令L[0] = a, (L[1]+...+L[n-1])/(n-1) = b
令f(k,n)是n条走廊，走过k条的状态
那么f(k,n)->f(k+1,n) 平均需要走(n-1)/(n-k)步(from Markov chain)
if(k!=1)，平均步长是 [a+(k-1)b]/k
if(k==1)，平均步长就是b
然后sum起来
这做法是不是笨了点？
【在 o****e 的大作中提到】
: pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
: 接。
: 我一道道加把
: 1。bionomial distribution近似
: http://www.mitbbs.com/article/Quant/31175297_0.html
: 这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
: np, then (n,k)*p^k*(1-p)^(n-
: k)--->lamda^k*exp(-lamda)/k!
: 可是这样还是心算不出来阿？ 大虾们说说把
: 2. random walk
| o****e
发帖数: 80 | 6 3rd question is same as coupon collection,
let xi be the additional pick up numbers for i th ball, then total pick up
numbers = x1+x2+..+x100,
E(X)=E(X1)+E(X2)+...E(X100)
after picked up i-1 balls already, the probabiliy to pick up any other balls
is p=[100-(i-1)]/100, and it is
geometric distribution, so the E[Xi]=1/p=100/[100-(i-1)],
so E(X)=100/100+100/99+..+100
who can explain problem 4?
【在 o****e 的大作中提到】
: pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
: 接。
: 我一道道加把
: 1。bionomial distribution近似
: http://www.mitbbs.com/article/Quant/31175297_0.html
: 这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
: np, then (n,k)*p^k*(1-p)^(n-
: k)--->lamda^k*exp(-lamda)/k!
: 可是这样还是心算不出来阿？ 大虾们说说把
: 2. random walk
| o****e
发帖数: 80 | 7 for problem 4, my answer is like this:
L1+2*L2+ (n-1)/(n-2)*2L3 + (n-1)/(n-3)*2L4 + ... +(n-1)/2*2Ln-1 + (n-1)*2Ln - Ln
=L1+2*L2+ 2*(n-1)*[L3/(n-2) + L4/(n-3) + ... + Ln-1/2 + Ln] - Ln
am I right?
【在 o****e 的大作中提到】
: pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
: 接。
: 我一道道加把
: 1。bionomial distribution近似
: http://www.mitbbs.com/article/Quant/31175297_0.html
: 这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
: np, then (n,k)*p^k*(1-p)^(n-
: k)--->lamda^k*exp(-lamda)/k!
: 可是这样还是心算不出来阿？ 大虾们说说把
: 2. random walk
| c**********s
发帖数: 295 | 8 ogtree, I think your solution is wrong. by symmetry the impact of L1, L2, …
, are all the same.
here is mine. it is in the same direction as native’s but I think he also
made some mistake in details.
again we have L0 = a, and sum(L1 to LN) / N = b
note in my set up I have totally N+1 path
first I move a, now I am at the center. from now on I do not count the run I
made to L0, we will consider it at the end.
so I have 2b * [1 + (N-1)/(N-1) + (N-1)/(N-2) + (N-1)/1] steps to cover all
the paths.
he | n****e
发帖数: 629 | 9 把N=3代进去试试？
I
all
【在 c**********s 的大作中提到】
: ogtree, I think your solution is wrong. by symmetry the impact of L1, L2, …
: , are all the same.
: here is mine. it is in the same direction as native’s but I think he also
: made some mistake in details.
: again we have L0 = a, and sum(L1 to LN) / N = b
: note in my set up I have totally N+1 path
: first I move a, now I am at the center. from now on I do not count the run I
: made to L0, we will consider it at the end.
: so I have 2b * [1 + (N-1)/(N-1) + (N-1)/(N-2) + (N-1)/1] steps to cover all
: the paths.
| c**********s
发帖数: 295 | 10 oh I missed one case in the last term. it can go from Li->L0->Li->L0...
the last term should be,
+ [2a/N + 2b/N^2 + 2a/N^3 + 2b /N^4 + ...] * [(N-1)/(N-1) + (N-1)/(N-2) + (N-1)/
1]
of course it can be simplified but I will leave it like this as it is more
intuitive. | | | c**********s
发帖数: 295 | 11 to summarise and correct some typos,
a
+ 2b * [1 + (N-1)/(N-1) + (N-1)/(N-2) + ... + (N-1)/1]
- b
+ [2a/N + 2b/N^2 + 2a/N^3 + 2b /N^4 + ...] * [(N-1)/(N-1) + (N-1)/(N-2) + ... +
(N-1)/1]
for [150,150,150]
a=150, b=150, N=2
150 + 300*2 -150 + 300*1 = 900 | c**********s
发帖数: 295 | 12 just read question 3, actually it is part of question 4 | b**********5
发帖数: 51 | 13 robustzgy:
1. use central limit theorem, the sum of 100 i.i.d. r.v. are close to
normal. let x=1 if head and -1 if tail, p=0.5, then sum(xi) ~ N(0, 100), P(
heads < 40) = P(sum of xi < -20) which is the left tail of 2*sigma, if you
still remembers the normal distribution the 2*sigma is about 2%, that's the
answer.
Questions:
1. Why the variance is 100?
2. Why P(heads < 40) = P(sum of xi < -20)?
Thanks. | r*******y
发帖数: 290 | 14 you should review some basic prob books
the
【在 b**********5 的大作中提到】
: robustzgy:
: 1. use central limit theorem, the sum of 100 i.i.d. r.v. are close to
: normal. let x=1 if head and -1 if tail, p=0.5, then sum(xi) ~ N(0, 100), P(
: heads < 40) = P(sum of xi < -20) which is the left tail of 2*sigma, if you
: still remembers the normal distribution the 2*sigma is about 2%, that's the
: answer.
: Questions:
: 1. Why the variance is 100?
: 2. Why P(heads < 40) = P(sum of xi < -20)?
: Thanks.
| b**********5
发帖数: 51 | 15 The variance of the sum of the independent variables is the sum of variances
of the independent variables given that they are independent variables.
Because we set x=1 if head and -1 if tail, p=0.5, so P(heads < 40) = P(sum
of xi < -20). | b**********5
发帖数: 51 | | b******e
发帖数: 118 | 17 For Q2, is it 1/3? But I'm not 100% sure.
P(T(1)
P(T(-1)
So answer is 0.5 * (2/3) = 1/3 | p*****k
发帖数: 318 | 18 bitalice, i think the answer to Q2 in the original thread is
correct. the only thing to note is when you calculate
Pr[T(-1)
(instead of 0), which results 1/3 instead of 2/3. | b******e
发帖数: 118 | 19 pcasnik, thanks a lot! You are right. | a********e
发帖数: 508 | 20 hey, your formula doesn't seem right if N=2
... +
【在 c**********s 的大作中提到】
: to summarise and correct some typos,
: a
: + 2b * [1 + (N-1)/(N-1) + (N-1)/(N-2) + ... + (N-1)/1]
: - b
: + [2a/N + 2b/N^2 + 2a/N^3 + 2b /N^4 + ...] * [(N-1)/(N-1) + (N-1)/(N-2) + ... +
: (N-1)/1]
: for [150,150,150]
: a=150, b=150, N=2
: 150 + 300*2 -150 + 300*1 = 900
| w*******x
发帖数: 489 | 21 第5题没有closed form
什么题目啊？
【在 o****e 的大作中提到】
: pcasnik大侠的建议，俺终于整明白了，他老人家让我看到有意思的老题目收集一下连
: 接。
: 我一道道加把
: 1。bionomial distribution近似
: http://www.mitbbs.com/article/Quant/31175297_0.html
: 这个题目，我只找到了 poisson 近似，when n big enough and p is small, lamda=
: np, then (n,k)*p^k*(1-p)^(n-
: k)--->lamda^k*exp(-lamda)/k!
: 可是这样还是心算不出来阿？ 大虾们说说把
: 2. random walk
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