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Transform of probability density from polar to cartesian

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Solution Preview

A vector is sampled from a probability distribution, such that the length of the vector has the probability density given by:

P(z)= zexp(-z^2/2) (1)

We assume that the angle of the vector with the positive x-direction is uniformly distributed between zero and 2 pi. Let's now define a probability density function P(z, theta) such that:

P(z, theta) dz d theta (2)

gives you the probability that z is between z and z + dz and theta is betwee theta and theta + d theta. We then must have that:

Integral over theta from theta = 0 to 2 pi of P(z, theta) d theta = P(z) --------------->

P(z, theta) = 1/(2 pi) P(z) = 1/(2 pi) z exp(-z^2/2) (3)

So, we see that the probability density as a function of z and theta is constant as a function of theta.

We now want to find the probability density of the cartesian coordinates (x,y) of the vector. Let Q(x, y) be this probability density. Then, by definition (see Appendix 1 for explanation), we have that:

Q(x, y)dx ...

Solution Summary

We give a detailed explanation of how to perform a general coordinate transformation on a probability distribution. We apply this to the given problem of transforming a probability distribution from polar to cartesian coordinates.

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