The calculation of the probability for the cross section of the joint distribution

Question

Given that I have a continuous joint distribution of two independent normal random variables (let the independent vars be on the X and Z axes, and the dependent one - the joint probability - on the Y axis), and I have a straight line somewhere on the XZ plane, like would I calculate the probability of a point falling from one or the other of this line?

Answer 1

First, move everything so that the two normal distributions (X and Z) are centered on zero; now the joint distribution will be a hill whose center is the origin.

, ( "" ). - .

, , - , . ( ), , Z. , X X . ( ), .

, .

EDIT: . ascii, .

, , sigmaX = sigmaZ = 1, :

joint probability: P(x, z) = 1/(2 pi) exp(-(x^2 + z^2)/2)

line: x = c

, , "" x x + dx:

P(x)dx = Int[z=-Inf, z=+Inf]{dz P(x, z)}
       = 1/sqrt(2 pi) exp(-x^2/2) 1/sqrt(2 pi) Int[z=-Inf, z=+Inf]{dz exp(-z^2/2)}
       = 1/sqrt(2 pi) exp(-x^2/2)

, () . , , , , ,

P(c>x) = Int[-Inf, c]{dx 1/sqrt(2 pi) exp(-x^2/2)}
       = 1/2 (1 - Erf(c/sqrt(2)))