Let (R,Θ) be a random vector with joint density f(r,θ)=1r2exp[−π(1+θ2)r] for r>0,θ∈R . What is the joint density of Φ(R,Θ) under the transformation Φ(r,θ)=(rcosθ,rsinθ) ?
Firstly we should check that the given function is a density. For each r , it is a Gaussian density whose integral is 1r2exp(−πr)1r√ .
We need to show that the integral is 1 . Putting u=πr we need to integrate 1π√u−−√exp(−u) which is 1 (gamma function integral).
Let x=rcos(θ) , y=rsin(θ) , then r2=x2+y2 and θ=tan−1(yx) .
So substitute these into f(r,θ) and multiply by the Jacobian: 1r .
However, we are not yet finished as the sine and cosine functions are periodic. We need to sum over all the solutions of tan(θ)=yx .
If the principal solution between −π2 and π2 is α then the density is 1r3exp(−πr)∑∞n=−∞exp(−(α+nπ)2)r) .
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Answer:
Let (R,Θ) be a random vector with joint density f(r,θ)=1r2exp[−π(1+θ2)r] for r>0,θ∈R . What is the joint density of Φ(R,Θ) under the transformation Φ(r,θ)=(rcosθ,rsinθ) ?
Firstly we should check that the given function is a density. For each r , it is a Gaussian density whose integral is 1r2exp(−πr)1r√ .
We need to show that the integral is 1 . Putting u=πr we need to integrate 1π√u−−√exp(−u) which is 1 (gamma function integral).
Let x=rcos(θ) , y=rsin(θ) , then r2=x2+y2 and θ=tan−1(yx) .
So substitute these into f(r,θ) and multiply by the Jacobian: 1r .
However, we are not yet finished as the sine and cosine functions are periodic. We need to sum over all the solutions of tan(θ)=yx .
If the principal solution between −π2 and π2 is α then the density is 1r3exp(−πr)∑∞n=−∞exp(−(α+nπ)2)r) .
It is neater to leave this in terms of r and α .