Fisher's Limiting spectral distribution

Question

Fisher's Limiting Spectral Distribution

Fs,t(x) = (1-t/2pi*(s+tx))sqrt((b-x)(x-a). where a is less than equal to x and x is less than equal to b.

a = (1-h)^2/(t-t^2), b= (1+h)^2/(1-t)^2 , h= (s+t-st)^1/2

a. Derive a formula for the first moment of the Fisher LSD in terms of its parameters s and t. you can assume that h>t as this always holds by the definition of h.

Answer 1

Hi Tanya,

To derive a formula for the first moment of Fisher's Limiting Spectral Distribution (LSD) in terms of its parameters s and t, we need to calculate the expected value of the random variable x under the probability distribution Fs,t(x).

The first moment, or expected value, is given by:

E[x] = ∫[a, b] x * Fs,t(x) dx

First, let's express Fs,t(x) in terms of its parameters:

Fs,t(x) = (1 - t/2π(s + tx)) * sqrt((b - x)(x - a))

Now, substitute the expressions for a, b, and h in terms of s and t:

a = (1 - h)^2 / (t - t^2) b = (1 + h)^2 / (1 - t)^2 h = √(s + t - st)

Next, let's calculate the expected value:

E[x] = ∫[a, b] x * (1 - t/2π(s + tx)) * sqrt((b - x)(x - a)) dx

This integral can be quite complex due to the square root terms. However, we can attempt to simplify it by making a substitution.

Let y = √((b - x)(x - a))

Now, dx = -2y dy / (b - a)

We also need to express x in terms of y:

x = (a + b - y^2) / 2

Now, we can rewrite the integral:

E[x] = ∫[y_min, y_max] [(a + b - y^2) / 2] * (1 - t/2π(s + t(a + b - y^2)/2)) * (-2y / (b - a)) dy

Where y_min and y_max are the values of y that correspond to x values a and b, respectively.

You can now compute this integral numerically. While it may not have a simple closed-form expression, you can use numerical methods or software to evaluate it for specific values of s and t.

This will give you the first moment of Fisher's Limiting Spectral Distribution in terms of the parameters s and t.

Hope this helps! If so, I would greatly appreciate your feedback as I am new to the platform.

Thank you,

Benjamin M.