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Quant版 - 问面试题
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相关话题的讨论汇总
话题: raining话题: rain话题: xy话题: predicts话题: 60%
进入Quant版参与讨论
1 (共1页)
i****b
发帖数: 52
1
When A predicts raining, the chance of raining is 60%. When B predicts
raining, the chance of raining is also 60%. If A and B both predict to rain
(assuming they are independent), what is the chance of raining?
s*********y
发帖数: 284
2
缺条件...
or you can make some assumption

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

i****b
发帖数: 52
3
nope. i asked whether the conditions are sufficient after 3 seconds and was
told yes....
h***s
发帖数: 2499
4
60%

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

l******o
发帖数: 28
5
我猜是0.72。假设了无条件下下雨概率是0.5
不确定是否正确 等待牛人给出详细解答
i****b
发帖数: 52
6
为啥是60%?。。。。 我说我只能是确定0.6 ~ 0.84之间。。 。 然后面试官让我继
续做。。。 到底也没做出来。。。。。

【在 h***s 的大作中提到】
: 60%
:
: rain

x***i
发帖数: 106
7
Stevens Capital Management?
我想假如没有下雨,就是both A and B are wrong,
If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16%
So P(rain) = 84%
我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以
最后结果应该是一个Range。
i****b
发帖数: 52
8
是这个公司。。。原来你也被问过阿哈。。 我觉得不能假设他们independent, 因为A
错了,说明不下雨,那么B一定错。所以0.84的上线是错的。 0.6的下限是对的。

【在 x***i 的大作中提到】
: Stevens Capital Management?
: 我想假如没有下雨,就是both A and B are wrong,
: If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16%
: So P(rain) = 84%
: 我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以
: 最后结果应该是一个Range。

x***i
发帖数: 106
9
恩,有道理。那么根据类似于Naive Bayes的原理,应该是这样的:
为了简洁,考虑条件概率下的Sample space:
Sample space given both predict "rain" (2 samples):
1. A correct and B correct
2. A incorrect and B incorrect
P(A correct and B correct | prior)= 0.6
P(A incorrect and B incorrect | prior) = 0.14
P(rain | both predict "rain") = P(A correct and B correct in the "codition"
sample space) = Bayes rule like Naive Bayes... = 0.81
我记得他提到过答案比0.84稍微小。

A

【在 i****b 的大作中提到】
: 是这个公司。。。原来你也被问过阿哈。。 我觉得不能假设他们independent, 因为A
: 错了,说明不下雨,那么B一定错。所以0.84的上线是错的。 0.6的下限是对的。

c**********6
发帖数: 18
10
为什么P(A incorrect and B incorrect | prior) = 0.14 ?难道不是 0.4?

"

【在 x***i 的大作中提到】
: 恩,有道理。那么根据类似于Naive Bayes的原理,应该是这样的:
: 为了简洁,考虑条件概率下的Sample space:
: Sample space given both predict "rain" (2 samples):
: 1. A correct and B correct
: 2. A incorrect and B incorrect
: P(A correct and B correct | prior)= 0.6
: P(A incorrect and B incorrect | prior) = 0.14
: P(rain | both predict "rain") = P(A correct and B correct in the "codition"
: sample space) = Bayes rule like Naive Bayes... = 0.81
: 我记得他提到过答案比0.84稍微小。

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进入Quant版参与讨论
r*****d
发帖数: 346
11
Since you mentioned independence, I will go for 84%. It is not because 84%
is The answer, but it is sufficient to make a case.

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

q*********i
发帖数: 21
12
有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B
predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y)
=0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=?
这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多
少都可以。

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

n******t
发帖数: 4406
13
我晕,金融界现在都这个level。

was

【在 i****b 的大作中提到】
: nope. i asked whether the conditions are sufficient after 3 seconds and was
: told yes....

N******K
发帖数: 10202
14
If A and B are two suckers, then nobody knows the chance of raining, could
be 0-1
If A and B are two gods, then 60% for sure
generally 60% +/- delta%
delta depends on the ability of the predictor

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

l******n
发帖数: 1250
15
我想这是一个boost的典型问题,multipredictor,how to sum up.其实值肯定大于0.6
, 但具体多少,很难确定
r*********n
发帖数: 4553
16
我觉得这个答案比较靠谱
看了下上面的答案,我觉得A,B对下雨的prediction和真正下雨的概率是没有关系的
因为A,B是独立的,所以A作为B的side information是没有价值的(反之亦然),所以
最后A,B的prediction都没有变化,还都是60% within certain confidence interval

could

【在 N******K 的大作中提到】
: If A and B are two suckers, then nobody knows the chance of raining, could
: be 0-1
: If A and B are two gods, then 60% for sure
: generally 60% +/- delta%
: delta depends on the ability of the predictor
:
: rain

T*****u
发帖数: 7103
17
60%吧,confidence interval变窄了而已;把a和b想像成一个coin摔两遍好了。
k******n
发帖数: 12
18

could
我认为是0.8, 原因如下:
当A和B都说下雨就真的下雨的概率均匀分布在0.6到1.0之间,均匀分布是因为他们相互
独立;在这种情况下,中间结果等于最终结果,即最后结果是(0.6+1.0)/2=0.8
可以看看A的概率和B的概率重叠的情况:
P(rain | A predicts rain)=0.6, 这个0.6可以叫0.6A
P(rain | B predicts rain)=0.6, 这个0.6可以叫0.6B
0.6A和0.6B重叠是x的情况下,得到P(rain | A and B both predict rain)=0.6+0.6-x
x=0.6时,0.6A和0.6B在重叠最多,得到P(rain | A and B both predict rain)=0.6
x=0.2时,0.6A和0.6B在重叠最少,得到P(rain | A and B both predict rain)=1.0
x在0.2到0.6之间均匀分布,x取0.2到0.6间任意值的概率是1/0.4,所以最后结果是
把(1.2-x)*1/0.4在0.2到0.6之间对x积分,最终得到0.8

【在 N******K 的大作中提到】
: If A and B are two suckers, then nobody knows the chance of raining, could
: be 0-1
: If A and B are two gods, then 60% for sure
: generally 60% +/- delta%
: delta depends on the ability of the predictor
:
: rain

K*V
发帖数: 192
19
Most of u are over thinking. I thought the answer is 0.6*0.6=0.36.

When A predicts raining, the chance of raining is 60%. When B predicts
raining, the chan........

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

l****w
发帖数: 583
20
我也认为是60%。
其实用土办法想想也很直观:A和B预测下雨的能力是一样的(就是6成的命中率)。他
们都一同预测下雨并不影响各自及最终的命中率。下雨概率和预测下雨的命中概率在这
个题目里好像整得挺晕的。
看了一下,回答60%的都是我喜欢的ID。哈哈。

interval

【在 r*********n 的大作中提到】
: 我觉得这个答案比较靠谱
: 看了下上面的答案,我觉得A,B对下雨的prediction和真正下雨的概率是没有关系的
: 因为A,B是独立的,所以A作为B的side information是没有价值的(反之亦然),所以
: 最后A,B的prediction都没有变化,还都是60% within certain confidence interval
:
: could

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hedge fund 都要什么背景的人?有人玩这个么? (转载)
只会Python 和 R,及predictive modeling能做矿工吗?Does non-stochastic mean predictable? Not necessary to be constant?
Paradox: B-S equation predicted stock price growth rate VS economy growth哪位大侠知道portfolio analysis一般都面试啥问题啊???
进入Quant版参与讨论
T*****u
发帖数: 7103
21
仁兄,如果问predict不下雨的概率捏?0.12? 那1-0.36-0.12=0.52 的概率属于啥?
predictor不工作概率?

【在 K*V 的大作中提到】
: Most of u are over thinking. I thought the answer is 0.6*0.6=0.36.
:
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chan........

c**********s
发帖数: 295
22
I seconde this.

B
Y)

【在 q*********i 的大作中提到】
: 有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B
: predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y)
: =0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=?
: 这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多
: 少都可以。
:
: rain

l***y
发帖数: 1166
23
P(AnB)=1-P(A'nB')
n means and
' means not
K*V
发帖数: 192
24
P(AUB) = 1-(A'nB')=1-(1-0.6)*(1-0.6) = 0.86

【在 l***y 的大作中提到】
: P(AnB)=1-P(A'nB')
: n means and
: ' means not

k*****h
发帖数: 12
25
认为答案是60%的同学,假如题目条件稍微改一下,改成A预测下雨,实际下雨的概率是
60%,B预测下雨,实际下雨的概率是80%,A B的预测互相独立,那么答案是?

【在 l****w 的大作中提到】
: 我也认为是60%。
: 其实用土办法想想也很直观:A和B预测下雨的能力是一样的(就是6成的命中率)。他
: 们都一同预测下雨并不影响各自及最终的命中率。下雨概率和预测下雨的命中概率在这
: 个题目里好像整得挺晕的。
: 看了一下,回答60%的都是我喜欢的ID。哈哈。
:
: interval

k*****h
发帖数: 12
26
second这位仁兄。个人觉得应该是这样求解,求指正:
已知:
P(Z|X) = 0.6
P(Z|Y) = 0.6
P(XY) = P(X)P(Y)
求P(Z|XY)
概率公式:
P(Z) = P(Z|X)P(X) (1)
P(Z) = P(Z|Y)P(Y) (2)
P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)
(1)*(2)/(3)得
P(Z|XY) = P(Z|X)*P(Z|Y)/P(Z)=0.6*0.6/P(Z)
P(Z)未知,所以无法求解。

B
Y)

【在 q*********i 的大作中提到】
: 有没有可能你听错题了,或者我理解错了。定义3个events, X=A predicts rain, Y=B
: predicts rain, Z=actually rain。根据你的描述,已知条件是:P(Z|X)=0.6, P(Z|Y)
: =0.6, P(XY)=P(X)P(Y),问题是:P(Z|XY)=?
: 这个答案从0到1都有可能,例子也不难构造。只要P(X),P(Y)足够小,P(Z|XY)想得多
: 少都可以。
:
: rain

k*****h
发帖数: 12
27
这位同学,你对P(A wrong)的理解不对
P(A wrong)有两种情况,一种是预报下雨却没下,一种是预报没下却下了。你只考虑了
一种情况。

【在 x***i 的大作中提到】
: Stevens Capital Management?
: 我想假如没有下雨,就是both A and B are wrong,
: If they are independent, P(A and B wrong) = P(A wrong)P(B wrong) = 16%
: So P(rain) = 84%
: 我记得没有给出independent的条件,所以如果A和B是完全正相关的话,会是60%。所以
: 最后结果应该是一个Range。

z****e
发帖数: 19
28
过程看起来没问题,不过最后结果假设p(z) = 0.1, p(z|xy) = 3.6?

【在 k*****h 的大作中提到】
: second这位仁兄。个人觉得应该是这样求解,求指正:
: 已知:
: P(Z|X) = 0.6
: P(Z|Y) = 0.6
: P(XY) = P(X)P(Y)
: 求P(Z|XY)
: 概率公式:
: P(Z) = P(Z|X)P(X) (1)
: P(Z) = P(Z|Y)P(Y) (2)
: P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)

a*******e
发帖数: 253
29
难道不是p(z,x)=p(z/x)p(x),还是我记错了?

【在 k*****h 的大作中提到】
: second这位仁兄。个人觉得应该是这样求解,求指正:
: 已知:
: P(Z|X) = 0.6
: P(Z|Y) = 0.6
: P(XY) = P(X)P(Y)
: 求P(Z|XY)
: 概率公式:
: P(Z) = P(Z|X)P(X) (1)
: P(Z) = P(Z|Y)P(Y) (2)
: P(Z) = P(Z|XY)P(XY) = P(Z|XY)P(X)P(Y) (3)

D********e
发帖数: 8
30
P(not rain) = P(A and B wrong) = P(A wrong)P(B wrong) = 16%
P(rain) = P(A and B right) = P(A right)P(B right) = 36%
The answer = 36/(36+16) = 36/52
Any comment?
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D********e
发帖数: 8
31
This problem is about given a and b give the same prediction, what is the
chance that they are both right to wrong.
The answer is clearly 50% if both prediction accuracy is 50%. So any
solution does not agree with special case won't be right.
If a 's accuracy is 50%, then the final answer should be the same as the b '
s prediction accuracy. Again any solution does not agree with this case won
't be right.
At least I. Am convinced that I am right.

【在 D********e 的大作中提到】
: P(not rain) = P(A and B wrong) = P(A wrong)P(B wrong) = 16%
: P(rain) = P(A and B right) = P(A right)P(B right) = 36%
: The answer = 36/(36+16) = 36/52
: Any comment?

H**********l
发帖数: 12
32
P(raining|A predicts raining)*P(raining|B predicts raining)/P(raining)

rain

【在 i****b 的大作中提到】
: When A predicts raining, the chance of raining is 60%. When B predicts
: raining, the chance of raining is also 60%. If A and B both predict to rain
: (assuming they are independent), what is the chance of raining?

1 (共1页)
进入Quant版参与讨论
相关主题
Does non-stochastic mean predictable? Not necessary to be constant?[合集] 请问牛人,下面说的是哪方面的东西。
哪位大侠知道portfolio analysis一般都面试啥问题啊???[合集] 中期底部
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Predictive Power of Hull White ModelThe next big boom, Do you know any heavy weight person who predict it ?
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相关话题的讨论汇总
话题: raining话题: rain话题: xy话题: predicts话题: 60%