Re: How to solve the probability density of z=ax+w?

On Mar 8, 6:34 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Mar 8, 5:41 pm, hhwolf76 <james.zho...@xxxxxxxxx> wrote:

This is a problem
If p(a=0)=1/2,p(a=1)=1/2
a,x,w are mutually independent. x,w are all normal distribution and their 1st and 2nd moment have been known. E(w)=0

Can I get the probability density of z=ax+w?

If a = 0 then the pdf of z is the same as the pdf of w, call it
P_0(z). If a = 1 then z = x + w, so z has a normal distribution whose
mean is the sum of the means of x and w, and whose variance is the sum
of the variances of x and w. Call this pdf P_1(z).

Then, unless I'm overlooking something, the overall pdf of z should be
1/2*(P_0(z) + P_1(z)) (which is, of course, *not* a normal
distribution).

and can I get E(x|z)?


As far as E(x|z) is concerned, the pdf of x given z, should, if you'll
excuse the sloppy notation, be given by

P(x|z) = Pr(x and z)/Pr(z)

where

Pr(x and z) = 1/2*N(mu_x, sigma_x, x)*N(mu_w, sigma_w, z - x) +
1/2*N(mu_x, sigma_x, x)*N(mu_w, sigma_w, z)

Pr(z) = 1/2*N(mu_xw, sigma_xw, z) + 1/2*N(mu_w, sigma_w, z)

and

N(mu, sigma, x) = 1/(sigma*sqrt(2*pi))*Exp(-(x - mu)^2/(2*sigma^2))
mu_x and sigma_x^2 are the mean and variance of x
mu_w and sigma_w^2 are the mean and variance of w
mu_xw = mu_x + mu_w and sigma_xw^2 = sigma_x^2 + sigma_w^2 are the
mean and variance of x + w

Then E(x|z) is found by integrating x*P(x|z) over x = -inf. to +inf.
Looking at it, I think that this can be done exactly in closed form.
The denominator Pr(z) is constant w.r.t x, and each of the products in
the numerator Pr(x and z) can be coerced into the functional form of a
normal distribution whose expected value is easily found. It "should
be" merely a question of slogging through a lot of algebra.

.

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