Harder brainteaser than the coin flip

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.

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.

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.