Probability distribution of two variables

[mjordan2nd]

Homework Statement

Consider a Hamiltonian involving two Gaussian variables, X and Y. Start from the statement that the average formed by these two variables is of the form

$$<e^{aX+bY}> = e^{a^2+b^2-ab}$$

Homework Equations

$$<e^{ax}> = \int_{-\infty}^{\infty} dx \frac{exp(-\beta ( x-<X>)^2/2)}{\sqrt{2 \pi / \beta}} e^{ax}$$

The Attempt at a Solution

Since X and Y are Gaussian the joint probability is P(x,y)=P(x)P(y) -- so we have two integrals. Using the definition above I found

$$<e^{ax}> = \sqrt{e^{\beta_x \left[ \left( <x> + \frac{a}{\beta_x} \right)^2 + <x> \right]}}$$
$$<e^{by}> = \sqrt{e^{\beta_y \left[ \left( <y> + \frac{b}{\beta_y} \right)^2 + <y> \right]}}$$

We can also say <exp[ax+by]> = <exp[ax]><exp[by]>, so combining the two above we find the desired expectation value. I'm not sure how to proceed from here to calculate the expectation values and the variances of the variables though. Any help would be appreciated. Thanks.

 

[mark726]

To calculate the expectation values and variances of X and Y, we can use the following equations:

< X > = -d/dx ln Z, where Z is the partition function

<V(x)> = <x^2> - <x>^2

Similarly for Y, we have:

< Y > = -d/dy ln Z

<V(y)> = <y^2> - <y>^2

To calculate the partition function Z, we can use the following equation:

Z = ∫∫ e^{-βH(x,y)} dx dy

Substituting the given Hamiltonian, we have:

Z = ∫∫ e^{-β(a^2+b^2-ab)} e^{-βX^2/2} e^{-βY^2/2} dx dy

= ∫ e^{-β(a^2+b^2-ab)} e^{-βX^2/2} dx ∫ e^{-βY^2/2} dy

= √(2π/β) √(2π/β) e^{β(a^2+b^2-ab)} e^{β<X>^2/2} e^{β<Y>^2/2}

= 2π/β e^{β(a^2+b^2-ab)} e^{β<X>^2/2} e^{β<Y>^2/2}

Now we can use this value of Z to calculate the expectation values and variances of X and Y using the equations mentioned above.