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I intended you to: find the joint density of X, Z by change of variables,
and then divide joint by marginal to get the conditional densities.
Here is one way to do part of the problem. Consider first the following
problem: U, V are bivariate normal with mean 0 and variance covariance
matrix [
1
]
ρ
ρ 1
Then
f U,V (u, v) =
1
2π √ − 2ρuv + v 2
1 − ρ 2 exp{−u2 }
2(1 − ρ 2 )
Rewrite u 2 − 2ρuv + v 2 as (u − ρv) 2 + (1 − ρ 2 )v 2 to see that
f V (v) =
1
2π √ 1 − ρ 2 e−v2 /2
∫ ∞
−∞
e −(u−ρv)2 /[2(1−ρ 2 )] du
In the integral substitute y = (u − ρv)/ √ 1 − ρ 2 and find that
f V (v) = 1 √
2π
e −v2 /2
The conditional density of U given V = v is then
f U|V (u|v) = f U,V (u, v)
f V (v)
which is the N(ρv, 1 − ρ 2 ) density.
√
(u − ρv)2
= exp{−
2(1 − ρ 2 ) − v2
2 + v2 2π
2 } 2π
= exp{−(u − ρv) 2 /[2(1 − ρ 2 )]}/ √ 2π(1 − ρ 2 )
Now if X = σU + µ and Z = φV + µ + γ then X, Z has the joint
distribution of X and X + Y of the original problem provided we make
ρ = Corr(X, Z). For X and Z as in the problem we find φ 2 = σ 2 + τ 2 ,
the covariance is σ 2 , and the correlation is
ρ =
1
√
1 + τ
2
/σ 2 =
σ
√
σ2 + τ 2 = σ φ
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