Question

Let \(X_{1}, X_{2}, X_{3}\) be iid random variables each having a standard normal distribution. Let the random variables \(Y_{1}, Y_{2}, Y_{3}\) be defined by $$ X_{1}=Y_{1} \cos Y_{2} \sin Y_{3}, \quad X_{2}=Y_{1} \sin Y_{2} \sin Y_{3}, \quad X_{3}=Y_{1} \cos Y_{3} $$ where \(0 \leq Y_{1}<\infty, 0 \leq Y_{2}<2 \pi, 0 \leq Y_{3} \leq \pi .\) Show that \(Y_{1}, Y_{2}, Y_{3}\) are mutually independent.

Short answer

The random variables \(Y_{1}, Y_{2}, Y_{3}\) are mutually independent as their joint pdf splits into the product of their marginal pdfs. Each \(Y_i\) has its own distribution. \(Y_1\) has a Rayleigh distribution, \(Y_2\) a uniform distribution on [0, 2\(\pi\)], and \(Y_3\) a cosine distribution on [0, \(\pi\)].

Step-by-step solution

Calculating the joint probability density function

First, calculate the joint probability density function of \(X_1\), \(X_2\), \(X_3\). By definition, this is given by the formula:\(f_{X_1,X_2,X_3}(x_1, x_2, x_3) = f_{X_1}(x_1) \cdot f_{X_2}(x_2) \cdot f_{X_3}(x_3)\), since \(X_1, X_2, X_3\) are independent. Given that each \(X_i\) has a standard normal distribution, this equals to: \(f_{X_1,X_2,X_3}(x_1, x_2, x_3) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x_1^2}{2}} \cdot \frac{1}{\sqrt{2\pi}}e^{-\frac{x_2^2}{2}} \cdot \frac{1}{\sqrt{2\pi}}e^{-\frac{x_3^2}{2}} = \frac{1}{(2\pi)^{3/2}}e^{-(x_1^2 + x_2^2 + x_3^2)/2}\)

Change of Variables for Joint Probability Distribution

The next step involves changing variables in the joint probability distribution from \(X_1, X_2, X_3\) to \(Y_1, Y_2, Y_3\). Use the Jacobian for transformations to get the absolute value of determinant of Jacobian matrix. Now, we have:\(\frac{\partial(X_1, X_2, X_3)}{\partial(Y_1, Y_2, Y_3)} = Y_1^2 \sin Y_3\). The absolute value of this determinant is the same, because it's already positive.

The Joint Distribution of Y Variables

To get the joint distribution of \(Y_1,Y_2,Y_3\), we plug all these into formula and simplify, taking notice of the bounds for \(Y_1, Y_2, Y_3\):\n\(f_{Y_1,Y_2,Y_3}(y_1, y_2, y_3) = f_{X_1,X_2,X_3}(x_1,y_2,y_3) \cdot• |\frac{\partial(X_1, X_2, X_3)}{\partial(Y_1, Y_2, Y_3)}| = \frac{1}{(2\pi)^{3/2}} \cdot e^{-\frac{Y_1^2}{2}}\cdot Y_1^2 \sin Y_3\)Given that \(0 \leq Y_1<\infty, 0 \leq Y_2<2 \pi, 0 \leq Y_3 \leq \pi\), it becomes evident that this is indeed the joint pdf for independent \(Y_1, Y_2, Y_3\) where each \(Y_i\) has its own distribution. \(Y_1\) has a Rayleigh distribution, \(Y_2\) a uniform distribution on [0, 2\(\pi\)], and \(Y_3\) a cosine distribution on [0, \(\pi\)]. Thus, we have shown that \(Y_1, Y_2, Y_3\) are mutually independent.