Expected value of |XY|^p for bivariate normal distribution where 0<p<2

README.md

Expected value of $|XY|^p$ for a bivariate normal distribution, where $(0 &lt; p &lt; 2)$

Disclaimer: This was generated using AI with Mathestral on Ollama and then edited by an underqualfied human.

Denote the bivariate normal random variables as $(X)$ and $(Y)$ with the following properties:

• Mean vector: $\mu = (\mu_X, \mu_Y)$

• Covariance matrix: $\Sigma = \binom{\sigma_X^2 , \rho \sigma_X \sigma_Y}{\rho \sigma_X \sigma_Y , \sigma_Y^2}$

The goal is to find $(\mathbb{E}[|XY|^p])$.

1. Transformation to standard bivariate normal:

   Assume $(X)$ and $(Y)$ are standard normal random variables with zero mean and unit variance with a correlation coefficient $(\rho)$.

2. Joint probability density function (PDF):

   The joint PDF of $(X)$ and $(Y)$ is given by:

   $f_{X,Y}(x, y) = \frac{1}{2\pi\sqrt{1 - \rho^2}} \exp \left( -\frac{1}{2(1 - \rho^2)} (x^2 - 2\rho xy + y^2) \right)$

3. Expectation calculation:

   $\mathbb{E}[|XY|^p] = \int_{-\infty}^\infty \int_{-\infty}^\infty |xy|^p f_{X,Y}(x, y) , dx , dy$

4. Simplifying the integral:

   Converting to polar coordinates where (x = r \cos\theta) and (y = r \sin\theta), and the Jacobian determinant is (r).

   The integral becomes:

   $\mathbb{E}[|XY|^p] = \int_0^\infty \int_0^{2\pi} |r^2 \cos\theta \sin\theta|^p \frac{r}{2\pi\sqrt{1 - \rho^2}} \exp \left( -\frac{r^2}{2(1 - \rho^2)}(1 - 2\rho \cos\theta \sin\theta) \right) r , d\theta , dr$

   Simplified:

   $\mathbb{E}[|XY|^p] = \frac{1}{2\pi\sqrt{1 - \rho^2}} \int_0^\infty r^{2p+1} \exp \left( -\frac{r^2}{2(1 - \rho^2)} \right) dr \int_0^{2\pi} |\cos\theta \sin\theta|^p \exp \left( \frac{r^2 \rho \cos\theta \sin\theta}{1 - \rho^2} \right) d\theta$

5. Separating the integrals: Considering the radial part and the angular part separately. The radial part is:

   $\int_0^\infty r^{2p+1} \exp \left( -\frac{r^2}{2(1 - \rho^2)} \right) dr$

   This is a gamma function integral:

   $\int_0^\infty r^{2p+1} e^{-\frac{r^2}{2(1 - \rho^2)}} dr = \left( \frac{2(1 - \rho^2)}{2} \right)^{p+1} \Gamma(p+1)$

   Simplified as:

   $(1 - \rho^2)^{p+1} \Gamma(p+1)$

6. Angular part: Assuming $(\rho = 0)$:

   $\int_0^{2\pi} |\cos\theta \sin\theta|^p d\theta = 2 \int_0^{\pi/2} (\cos\theta \sin\theta)^p d\theta = 2 \int_0^{\pi/2} \cos^p \theta \sin^p \theta d\theta$

   Using the Beta function $(B(x,y))$:

   $2 \int_0^{\pi/2} \cos^p \theta \sin^p \theta d\theta = 2 \cdot B\left( \frac{p+1}{2}, \frac{p+1}{2} \right) = 2 \frac{\Gamma(\frac{p+1}{2}) \Gamma(\frac{p+1}{2})}{\Gamma(p+1)}$

7. Combining results:

   $\mathbb{E}[|XY|^p] = \frac{(1 - \rho^2)^{p+1} \Gamma(p+1)}{2\pi\sqrt{1 - \rho^2}} 2 \frac{\Gamma(\frac{p+1}{2})^2}{\Gamma(p+1)}$

   Simplified, the terms involving $(\Gamma(p+1))$ cancel:

   $\mathbb{E}[|XY|^p] = \frac{(1 - \rho^2)^{p+1}}{\sqrt{1 - \rho^2}} \frac{\Gamma(\frac{p+1}{2})^2}{\pi}$

Therefore, the expected value $\mathbb{E}[|XY|^p]$ for $(0 &lt; p &lt; 2)$ is:

$\boxed{\mathbb{E}[|XY|^p] = (1 - \rho^2)^p \frac{\Gamma\left(\frac{p+1}{2}\right)^2}{\pi}}$

---

R Function

expected_value_XY_p <- function(p, rho) {
  if (p <= 0 || p >= 2) {stop("p must be between 0 and 2.")}
  
  ## Compute the Gamma function values
  gamma_half_p1 <- gamma((p + 1) / 2)
  
  ## Compute the expected value
  expected_value <- (1 - rho^2)^p * (gamma_half_p1^2) / pi
  
  return(expected_value)
}

## Example usage
p <- 1.5
rho <- 0.5
result <- expected_value_XY_p(p, rho)
print(result)
# [1] 0.1698573

Disclaimer: This was generated using AI with Mathestral on Ollama and then edited by an underqualfied human.