The Bivariate Cauchy Probability Density Function f is defined over the whole plane D = R² by 1 f(x, y) 1 1+ x² + y²
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![## Definition 1: The Bivariate Cauchy Probability Density Function
The bivariate Cauchy probability density function \( f \) is defined over the whole plane \( \mathcal{D} = \mathbb{R}^2 \) by:
\[
f(x, y) = \frac{1}{\pi} \cdot \left( \frac{1}{1 + x^2 + y^2} \right)^2
\]
## Problem 1
### (1.1) Verification of Probability Density Function
To verify that \( f \) is a probability density function, calculate:
\[
I := \iint_{\mathbb{R}^2} \frac{1}{\pi} \cdot \left( \frac{1}{1 + x^2 + y^2} \right)^2 \, dA
\]
### (1.2) Locating the Mean
To locate the mean \( (\bar{x}, \bar{y}) \) of \( f \), calculate or determine by mathematical considerations:
\[
\bar{x} := \frac{1}{I} \iint_{\mathbb{R}^2} x \cdot \frac{1}{\pi} \cdot \left( \frac{1}{1 + x^2 + y^2} \right)^2 \, dA
\]
\[
\bar{y} := \frac{1}{I} \iint_{\mathbb{R}^2} y \cdot \frac{1}{\pi} \cdot \left( \frac{1}{1 + x^2 + y^2} \right)^2 \, dA
\]
### (1.3) Determining the Existence of Variance
To determine whether the variance exists, calculate:
\[
(\bar{x} - x)^2 := \frac{1}{I} \iint_{\mathbb{R}^2} (x - \bar{x})^2 \cdot \frac{1}{\pi} \cdot \left( \frac{1}{1 + x^2 + y^2} \right)^2 \, dA
\]
\[
(\bar{y} - y)^2 := \frac{1}{I} \iint_{\mathbb{R}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2F53254225-3209-4514-903e-a417b44a0a3d%2Fw59fmob_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 5 images
