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




I just don't know how do you get this Pr(x and z) ?






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

and

N(mu, sigma, x) = 1/(sigma*sqrt(2*pi))*Exp(-(x -
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 +
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.


--
Hui
.

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