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Getting the joint probability density for the characteristic equation

  1. Dec 3, 2013 #1
    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: [itex]f_z (k) = <e^{ikz}> = \int_{-\infty}^{+\infty} e^{ikz}\, P_z (z) dz[/itex]

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

    Also, [itex]P_{x,y} = P_x (x) \, P_y (y) [/itex] for independent stochastic variables x and y.

    For Gaussian distribution: [itex] P_x = \frac{1}{\sqrt{2∏} } e^{\frac{-x^2}{2}} [/itex]

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

    Since [itex] G(x,y)= x^2 +y^2 [/itex] and [itex]P_{x,y} = P_x (x) \, P_y (y) [/itex]

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

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

    [itex] 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) [/itex]

    [itex] 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) [/itex]

    [itex] 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) [/itex]

    [itex] 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) [/itex]

    [itex] 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) [/itex]

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

    Thank you very much for your help! :biggrin:
     
  2. jcsd
  3. Dec 3, 2013 #2

    tiny-tim

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    hi physicsjn! :smile:
    change to ∫∫ rdrdθ, and then it's just δ(z - r2) :wink:
     
  4. Dec 3, 2013 #3
    Are the integration limits correct?

    Thank you tiny-tim! :biggrin: So I guess this will become
    [itex]\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)[/itex]

    [itex] = 2\pi\int^{+\infty}_{-\infty} r \, dr \, \, \delta(z-r^2)[/itex]

    [itex] = 2\pi \, \frac{r^2}{2} \, \delta(z-r^2)\,|^{+\infty}_{-\infty} [/itex]

    [itex] = \pi \, r^2 \, \delta(z-r^2)\,|^{+\infty}_{-\infty} [/itex]

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

    leaving us with

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

    Is this correct? :shy: Thanks again.. :)
     
  5. Dec 4, 2013 #4

    tiny-tim

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    hi physicsjn! :smile:

    (just got up :zzz:)

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

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

    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 :confused:)
     
  6. Dec 5, 2013 #5
    Thank you tiny-tim! :biggrin:

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

    And solving with this new limits should yield
    [itex]\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)[/itex]

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

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

    But here's what I'm trying to say:
    We want to evaluate [itex]2\pi\int^{+\infty}_{0} r \, dr \,δ(z-r^2)[/itex]
    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 [itex]r^2=z[/itex] or when [itex]r=\sqrt{z}[/itex]. 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:[itex]r=\sqrt{z}[/itex]. The rest of the terms are just zeroes because of delta function.
    So I thought [itex] \int^{+\infty}_{0} r \, dr \,δ(z-r^2)=\sqrt{z} [/itex]
    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. :smile:

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

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