17.2. Independence

Jointly distributed random variables \(X\) and \(Y\) are independent if \($ P(X \in A, Y \in B) = P(X \in A)P(Y \in B) $\) for all intervals \(A\) and \(B\).

Let \(X\) have density \(f_X\), let \(Y\) have density \(f_Y\), and suppose \(X\) and \(Y\) are independent. Then if \(f\) is the joint density of \(X\) and \(Y\),

\[\begin{split} \begin{align*} f(x, y)dxdy &\sim P(X \in dx, Y \in dy) \\ &= P(X \in dx)P(Y \in dy) ~~~~~ \text{(independence)} \\ &= f_X(x)dx f_Y(y)dy \\ &= f_X(x)f_Y(y)dxdy \end{align*} \end{split}\]

Thus if \(X\) and \(Y\) are independent then their joint density is given by

\[ f(x, y) = f_X(x)f_Y(y) \]

This is the product rule for densities: the joint density of two independent random variables is the product of their densities.

17.2.1. Independent Standard Normal Random Variables

Suppose \(X\) and \(Y\) are i.i.d. standard normal random variables. Then their joint density is given by

\[ f(x, y) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}y^2}, ~~~~ -\infty < x, y < \infty \]

Equivalently, \($ f(x, y) = \frac{1}{2\pi} e^{-\frac{1}{2}(x^2 + y^2)}, ~~~~ -\infty < x, y < \infty $\)

Here is a graph of the joint density surface.

def indep_standard_normals(x,y):
    return 1/(2*math.pi) * np.exp(-0.5*(x**2 + y**2))

Plot_3d((-4, 4), (-4, 4), indep_standard_normals, rstride=4, cstride=4)

Notice the circular symmetry of the surface. This is because the formula for the joint density involves the pair \((x, y)\) through the expression \(x^2 + y^2\) which is symmetric in \(x\) and \(y\).

Notice also that \(P(X = Y) = 0\), as the probability is the volume over a line. This is true of all pairs of independent random variables with a joint density: \(P(X = Y) = 0\). So for example \(P(X > Y) = P(X \ge Y)\). You don’t have to worry about whether or not to the inequality should be strict.

17.2.2. The Larger of Two Independent Exponential Random Variables

Let \(X\) and \(Y\) be independent random variables. Suppose \(X\) has the exponential \((\lambda)\) distribution and \(Y\) has the exponential \((\mu)\) distribution. The goal of this example is to find \(P(Y > X)\).

By the product rule, the joint density of \(X\) and \(Y\) is given by

\[ f(x, y) ~ = ~ \lambda e^{-\lambda x} \mu e^{-\mu y}, ~~~~ x > 0, ~ y > 0 \]

The graph below shows the joint density surface in the case \(\lambda = 0.5\) and \(\mu = 0.25\), so that \(E(X) = 2\) and \(E(Y) = 4\).

def independent_exp(x, y):
    return 0.5 * 0.25 * np.e**(-0.5*x - 0.25*y)

Plot_3d((0, 10), (0, 10), independent_exp)

To find \(P(Y > X)\) we must integrate the joint density over the upper triangle of the first quadrant, a portion of which is shown below.


The probability is therefore \($ P(Y > X) ~ = ~ \int_0^\infty \int_x^\infty \lambda e^{-\lambda x} \mu e^{-\mu y} dy dx $\)

We can do this double integral without much calculus, just by using probability facts.

\[\begin{split} \begin{align*} P(Y > X) &= \int_0^\infty \int_x^\infty \lambda e^{-\lambda x} \mu e^{-\mu y} dy dx \\ \\ &= \int_0^\infty \lambda e^{-\lambda x} \big{(} \int_x^\infty \mu e^{-\mu y} dy\big{)} dx \\ \\ &= \int_0^\infty \lambda e^{-\lambda x} e^{-\mu x} dx ~~~~~~ \text{(survival function of } Y\text{, evaluated at } x \text{)} \\ \\ &= \frac{\lambda}{\lambda + \mu} \int_0^\infty (\lambda + \mu) e^{-(\lambda + \mu)x} dx \\ \\ &= \frac{\lambda}{\lambda + \mu} ~~~~~~~ \text{(total integral of exponential } (\lambda + \mu) \text{ density is 1)} \end{align*} \end{split}\]


\[ P(Y > X) ~ = ~ \frac{\lambda}{\lambda + \mu} \]


\[ P(X > Y) ~ = ~ \frac{\mu}{\lambda + \mu} \]

Notice that the two chances are proportional to the parameters. This is consistent with intuition if you think of \(X\) and \(Y\) as two lifetimes. If \(\lambda\) is large, the corresponding lifetime \(X\) is likely to be short, and therefore \(Y\) is likely to be larger than \(X\) as the formula implies.

If \(\lambda = \mu\) then \(P(Y > X) = 1/2\) which you can see by symmetry since \(P(X = Y) = 0\).

If we had attempted the double integral in the other order – first \(x\), then \(y\) – we would have had to do more work. The integral is

\[ \int_0^\infty \int_0^y \lambda e^{-\lambda x} \mu e^{-\mu y} dx dy \]

Let’s take the easy way out by using SymPy to confirm that we will get the same answer.

# Create the symbols; they are all positive

x = Symbol('x', positive=True)
y = Symbol('y', positive=True)
lamda = Symbol('lamda', positive=True)
mu = Symbol('mu', positive=True)
# Construct the expression for the joint density

f_X = lamda * exp(-lamda * x)
f_Y = mu * exp(-mu * y)
joint_density = f_X * f_Y
# Display the integral – first x, then y

Integral(joint_density, (x, 0, y), (y, 0, oo))
# Evaluate the integral

answer = Integral(joint_density, (x, 0, y), (y, 0, oo)).doit()
# Confirm that it is the same 
# as what we got by integrating in the other order