# Getting the joint probability density for the characteristic equation

1. Dec 3, 2013

### physicsjn

Dear all,

Greetings! I was given a problem from Reichl's Statistical Physics book. Thank you very much for taking time to read my post.

1. The problem statement, all variables and given/known data

The stochastic variables X and Y are independent and Gaussian distributed with
first moment <x> = <y> = 0 and standard deviation σx = σy = 1. Find the characteristic function
for the random variable Z = X2+Y2, and compute the moments <z>, <z2> and <z3>. Find the first 3 cumulants.

2. Relevant equations
Characteristic equation: $f_z (k) = <e^{ikz}> = \int_{-\infty}^{+\infty} e^{ikz}\, P_z (z) dz$

Joint Probability density: $P_z(z) = \int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty} dy \, δ (z - G(x,y)) P_{x,y}(x,y)$ where $z = G (x, y)$

Also, $P_{x,y} = P_x (x) \, P_y (y)$ for independent stochastic variables x and y.

For Gaussian distribution: $P_x = \frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}}$

3. The attempt at a solution
To get the characteristic equation, we need first to get the joint probability density Pz(z):

Since $G(x,y)= x^2 +y^2$ and $P_{x,y} = P_x (x) \, P_y (y)$

$P_z(z) = \int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty} dy \, δ (z - x^2 +y^2) P_x (x) P_y (y)$

$P_z(z) = \int_{-\infty}^{+\infty}P_x (x) \, dx \, \int_{-\infty}^{+\infty}P_y (y) \, dy \, δ (z - x^2 +y^2)$

$P_z(z) = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}} \, dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-y^2}{2}} \, dy \, δ (z - x^2 +y^2)$

$P_z(z) = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}} \, dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-1}{2}(z-x^2)} \, dy \, δ (z - x^2 +y^2)$

$P_z(z) = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-1}{2}(x^2+z-x^2)} \, dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } dy \, δ (z - x^2 +y^2)$

$P_z(z) = \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } e^{\frac{-1}{2}z} \, dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } dy \, δ (z - x^2 +y^2)$

$P_z(z) = \frac{1}{\sqrt{2∏} } e^{\frac{-1}{2}z} \,\int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } dy \, δ (z - x^2 +y^2)$

Question: How do I simplify this factor $\int_{-\infty}^{+\infty} dx \, \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2∏} } dy \, δ (z - x^2 +y^2)$ ?

Thank you very much for your help!

2. Dec 3, 2013

### tiny-tim

hi physicsjn!
change to ∫∫ rdrdθ, and then it's just δ(z - r2)

3. Dec 3, 2013

### physicsjn

Are the integration limits correct?

Thank you tiny-tim! So I guess this will become
$\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}dx \,dy \, \delta(z-x^2-y^2) = \int^{+\infty}_{-\infty}\int^{2\pi}_{0} r \, dr \, d\theta \, \delta(z-r^2)$

$= 2\pi\int^{+\infty}_{-\infty} r \, dr \, \, \delta(z-r^2)$

$= 2\pi \, \frac{r^2}{2} \, \delta(z-r^2)\,|^{+\infty}_{-\infty}$

$= \pi \, r^2 \, \delta(z-r^2)\,|^{+\infty}_{-\infty}$

Now, because of the delta term, all other r will be killed except when r2=z,

leaving us with

$\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}dx \,dy \, \delta(z-x^2-y^2) = \pi \, z$

Is this correct? :shy: Thanks again.. :)

4. Dec 4, 2013

### tiny-tim

hi physicsjn!

(just got up :zzz:)

i] $\int^{+\infty}_{-\infty}\int^{2\pi}_{0}$ covers the plane twice, doesn't it?

ii]
i don't follow what you're doing here

i'd say (for z > 0) ∫0 r δ(z - r2) dr

= ∫0 1/2 δ(z - u) du

= 1/2

(not sure that works for z = 0 )

5. Dec 5, 2013

### physicsjn

Thank you tiny-tim!

i]
Ahhh... I see... My limits should be
$\int^{+\infty}_{0} \int^{2\pi}_{0}$

And solving with this new limits should yield
$\int^{+\infty}_{-\infty} \int^{+\infty}_{-\infty} dx \, dy \, δ(z-x^2-y^2)=\int^{+\infty}_{0} \int^{2\pi}_{0}r \, dr \, d\theta \, δ(z-r^2)$

$=2\pi\int^{+\infty}_{0} r \, dr \,δ(z-r^2)$

ii]
I'm really sorry. The second line $=2\pi\frac{r^2}{2}\delta(z-r^2) |_{-\infty}^{+\infty}$is wrong. Just forget that I have written it.

But here's what I'm trying to say:
We want to evaluate $2\pi\int^{+\infty}_{0} r \, dr \,δ(z-r^2)$
In my mind, I picture this integral as a sum of all the possible r values from 0 to infinity. However, the delta factor is zero everywhere except when $r^2=z$ or when $r=\sqrt{z}$. In this case the delta function is just equal to one.
Hence, despite the integral being an infinite sum of r's, only one term will survive:$r=\sqrt{z}$. The rest of the terms are just zeroes because of delta function.
So I thought $\int^{+\infty}_{0} r \, dr \,δ(z-r^2)=\sqrt{z}$
It's similar to the what I remember as Fourier's trick. But I guess using your relation below makes things simpler, so just forget this Fourier's trick stuff.

[iii]
Wow! I totally don't know this. Thank you very much for this! I tried to show this in my to-be-submitted solution. I let u=r2 and solved for du and found $\frac{du}{2}=r \, dr$. I can't see though why it shouldn't work for z=0. I'll just assume it'll work.

I think from here the rest is just plug and chug. Thank you so much tiny-tim for helping me! Sorry if I'm a bit slow.