Suppose X ~ Normal(0; 1) and Y = X + epsilon, where epsilon ~ Normal(0; sigma^2) and epsilon is independent of X. Conditional on observing Y , what is the distribution of X?
SBs for 1st correct answer
Edit - this is in response to //www.wallstreetoasis.com/forums/the-hardest-brainteaser-that-you-ever-did-see
Edit - solved below so scrolling down is cheating if you want to solve.
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Well, by my calculations, the correct answer is -theta --- +epsilon(X-Y^epsilon+1)
Well, by my calculations, the correct answer is -theta --- +epsilon(X-Y^epsilon+1)
Bivariate Normal Distribution FTW
Zy,Zx f(x, y)dxdy = 1 f(x, y) >= 0
Where Z is the integral
the distribution of X is the same as the distribution of Y.
Y = X + epsilon, so X = Y - epsilon. Since epsilon is an unbiased normal variate, X and Y are equivalent under permutation. By symmetry, X and Y follow the same distribution.
lol
then i will go with my second choice: X ~ Normal(0; 1).
i hope this isn't a gag question.
11
X~N(y, sigma^2)
where Y observed is Y=y
This puzzle cannot be solved due to chaos theory . . .
normal (Y, sigma^2) ?
What kind of interview are they asking you this in?
Hey,
Is the answer:
P(X|Y) ~ N(rho*y/(sigma^2+1),(1-rho)^2)
where rho is the correlation coefficient of X and Y? (i.e. rho = Cov(X,Y)/(Var(X)Var(Y)) = Cov(X,Y)/(sigma^2+1))
I had to use paper and pencil for this [x has been edited to y, since y is observed...]
Ok what about this:
P(X|Y) ~ N(rho*y/A, 1-rho^2)
where A = sqrt(sigma^2+1)
I spent some time on this now, is it wrong? Again rho is the correlation coeff. AFAIK all Gaussians used here should be 1-D.
The answer is 42.
Can you pm me the answer? I'm kinda dying here.
Your answer is looking better. Hint - there is no rho in the answer. Let me know if you still want the answer.
Dr ... this isn't that hard.
(X|Y) ~ N(Y/(1+sig^2), (sig^2)/(1+sig^2))
Also ML, you can compute rho explicitly given independence of X and eps...you know the formula for covariance and you know the variances of X and Y....
Also, I never said this was that hard. But I think it is a very good question to test overall comfort with conditional probability. Just having the insight to say that E[X|Y] =!= Y, and E[X|Y] is between 0 and Y is sufficient here for most purposes.
Shouldnt the E(X|Y)=Y/(1+sigma) without the square power?
Also, a note to the above who fell into the trick of thinking the mean was Y (which is a gut instinct), an easy demonstration of the fallacy occurs when we allow
Y = X1 + X2, with X1 and X2 iid. (this is generalizable to Y = Sigma(X_i))
Then, clearly, E(X1|Y) = E(X2|Y) and by linearity E(X1|Y) + E(X2|Y) = Y, so E(X1|Y) = Y/2.
A more general form of this reasoning is used to give the result here.
Ok, I guessed I rushed it, yes rho = 1/sqrt(1+sigma^2) with X and epsilon independent. I don't know if this counts as a brainteaser though - its not involved but I doubt this will be asked at general analysts' interview, or maybe I'm wrong.
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