Final answer:
The joint density function of U=X and V=X/Y, when X and Y are independent standard normal variables, is found by multiplying their individual densities. Further to this, by integrating the joint density function over all X while treating V as constant, it can be shown that V=X/Y has a Cauchy distribution.
Step-by-step explanation:
To determine the joint density function of U = X and V = X/Y when X and Y are independent standard normal random variables, we first recall that the standard normal distribution for a continuous random variable X is denoted as X ~ N(0, 1). Since U = X, the density function of U is just that of X, which is the standard normal density function. For V = X/Y, because the variables are independent, we can use the transformation technique to find the joint density function.
The probability density function of a standard normal variable is given by φ(x) = 1/√(2π) * exp(-x^2/2). Since X and Y are independent, their joint density function is the product of their marginal densities: f(x, y) = φ(x) * φ(y). For V, we let W = 1/Y, thus the density function for W when Y is standard normal, is also standard normal since the reciprocal of a standard normal variable is also a standard normal, albeit over a different domain (y ≠ 0).
Next, we note that the random variable X/Y has a Cauchy distribution, a well-known result in probability theory. This is shown by finding the distribution function of V by integrating the joint density function over the entire range of X, treating V as a constant. The result of this integration yields the density function of a Cauchy distribution, highlighting that X/Y indeed follows a Cauchy distribution with parameters (0,1).