Solving Joint Convolution for P(X,Y) - 65 Characters

  • Thread starter Thread starter yamdizzle
  • Start date Start date
  • Tags Tags
    Convolution Joint
yamdizzle
Messages
14
Reaction score
0
I solved majority of the question I just need to find the last joint density. Found the equations at part 3.

Homework Statement



Show P(X-Y=z ,Y=y) = P(X) = P(|Y|)
I showed P(X) = P(|Y|)

Homework Equations

The Attempt at a Solution


P(X=x,Y=y) = \frac{2*(2x-y)}{\sqrt{2πT^3σ^6}} * exp(((\frac{-(2x-y)^2}{(2σ^2T)}))
P(Y=y) = NormalPDF(0,Tσ^2)
P(X=x) = 2*NormalPDF(0,Tσ^2)

I don't really want to find the convolution then the Jacobian unless I have to. If there is an easier way please let me know.
 
Last edited:
Physics news on Phys.org
I took a step ahead and said:

P(X=x, Y=y) = P(X-Y=x-y , Y=y) = P(Z=z, Y=y) but I don't seem to get the right distribution.
 
yamdizzle said:
I solved majority of the question I just need to find the last joint density. Found the equations at part 3.

Homework Statement



Show P(X-Y=z ,Y=y) = P(X) = P(|Y|)
I showed P(X) = P(|Y|)

Homework Equations




The Attempt at a Solution


P(X=x,Y=y) = \frac{2*(2x-y)}{\sqrt{2πT^3σ^6}} * exp(((\frac{-(2x-y)^2}{(2σ^2T)}))
P(Y=y) = NormalPDF(0,Tσ^2)
P(X=x) = 2*NormalPDF(0,Tσ^2)

I don't really want to find the convolution then the Jacobian unless I have to. If there is an easier way please let me know.

Ignore the question There was a typo at the question. I just solved it Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top