3AM stats quiz

PhilGoetz

Because I'm tired and easily amused.

Consider two independent, normally-distributed variables X and Y, each with a mean of 0.
- X has a standard deviation of 4.
- Y has a standard deviation of 1.
Consider the probability distribution function (in two dimensions) for <X, Y>.
Imagine contour lines on the PDF connecting points of equal probability density. These would be a series of concentric ellipses.
Draw the contour line that is tangent to the line X = 4.
What values of Y does it pass through when X = 0?
Post your answer.
Run this R code to plot the contours:

library(mvtnorm)

bivn <- mvrnorm(100000, mu = c(0, 0), Sigma = matrix(c(4, 0, 0, 1), 2, 2))

bivn.kde <- kde2d(bivn[,1], bivn[,2], n = 100)

image(bivn.kde); contour(bivn.kde, add = T)

Because I'm tired and easily amused.

Consider two independent, normally-distributed variables X and Y, each with a mean of 0.
- X has a standard deviation of 4.
- Y has a standard deviation of 1.
Consider the probability distribution function (in two dimensions) for <X, Y>.
Imagine contour lines on the PDF connecting points of equal probability density. These would be a series of concentric ellipses.
Draw the contour line that is tangent to the line X = 4.
What values of Y does it pass through when X = 0?
Post your answer.
Run this R code to plot the contours:

library(mvtnorm)

bivn <- mvrnorm(100000, mu = c(0, 0), Sigma = matrix(c(4, 0, 0, 1), 2, 2))

bivn.kde <- kde2d(bivn[,1], bivn[,2], n = 100)

image(bivn.kde); contour(bivn.kde, add = T)

Used rot13 to avoid spoilers:

K unf qrafvgl N rkc(-k^2 / 32) jurer N = 1/(4 fdeg(2*cv))

L unf qrafvgl O rkc(-l^2 / 2) jurer O = 1/fdeg(2cv)

Fvapr gurl'er vaqrcraqrag gur wbvag qrafvgl vf gur cebqhpg bs gur vaqvivqhny qrafvgvrf, anzryl NO * rkc(-k^2 / 32 - l^2 / 2). Urapr, gur pbagbhe yvarf fngvfsl -k^2 / 32 - l^2 / 2 = pbafgnag. Nofbeovat gur artngvir vagb gur pbafgnag, jr trg k^2 / 32 + l^2 / 2 = pbafgnag, juvpu vf na ryyvcfr jvgu nkrf cnenyyry gb gur pbbeqvangr nkrf.

Guvf vf gnatrag gb gur yvar K = 4 ng gur cbvag (4,0). Fhofgvghgvat vagb gur rdhngvba sbe gur ryyvcfr, jr svaq gung gur pbafgnag vf 1/2, fb k^2 / 32 + l^2 / 2 = 1/2. Frggvat k = 0 naq fbyivat sbe l, jr svaq gung l = 1.

Small quibble: +- at the end.

[EDITED to be less spoilery.]

0PhilGoetz11y

You're right. That's what I expected, but not the answer I was getting. I was plotting bivariate distributions, it was 3AM, and a parameter that I thought was a standard deviation was a variance.

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