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Unformatted text preview: e second factor is the conditional pdf of Y given X = u :
√ exp −
2π fX (u) = 1 fY |X (v |u) = 2π (1 − ρ2 ) exp − (v − ρu)2
2(1 − ρ2 ) . (4.39) Thus, X is a standard normal random variable. By symmetry, Y is also a standard normal random
variable. This proves (a).
The class of bivariate normal pdfs is preserved under linear transformations corresponding to
multiplication of X by a matrix A if det A = 0. Given a and b, we can select c and d so that
the matrix A = a d has det(A) = ad − bc = 0. Then the random vector A X has a bivariate
normal pdf, so by part (a) already proven, both of its coordinates are Gaussian random variables.
In particular, its ﬁrst coordinate, aX + bY, is a Gaussian random variable. This proves (b).
By (4.39), given X = u, the conditional distribution of Y is Gaussian with mean ρu and variance
1 − ρ2 . Therefore, g ∗ (u) = E [Y |X = u] = ρu. Since X and Y are both standard (i.e. they have
mean zero and variance one), ρX,Y...
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- Spring '08
- The Land