Continuous Random Variable

Study guides: Exam P-1, Exam ASTAM

A Continuous Random Variable X is a Random Variable whose CDF F(x) is the integral of a density f(x), so that probability is area under f and every single value has probability zero.

F(x) = P(X \le x) = \int_{-\infty}^{x} f(t)\,dt

f(x) = F'(x)

• A valid density has f(x) \ge 0 and \int_{-\infty}^{\infty} f(x)\,dx = 1. It is a rate, not a probability, so it can exceed 1: \text{Uniform}(0, 0.5) has f(x) = 2. The named families are in Continuous Univariate Distributions.
• P(X = c) = 0 for every c, so P(a < X < b) = P(a \le X \le b) = F(b) - F(a). Strict and weak inequalities give the same answer — unlike a Discrete Random Variable.
• Expectations replace sums with integrals: E[g(X)] = \int g(x)\,f(x)\,dx (see Expected Value and Moment). For X \ge 0 the survival shortcut E[X] = \int_0^\infty [1 - F(x)]\,dx often avoids integration by parts.
• With two or more continuous variables the same calculations run on a Joint Probability Density Function: probabilities, marginals and moments become double integrals, and the work is setting up the Region of Integration.
• A linear combination of independent normal variables is exactly normal — the continuous case Exam P singles out in Probabilities for Linear Combinations.
• Insurance payments are often mixed. A continuous loss passed through a Deductible or a limit produces a payment with a point mass at 0 or at the limit, which is neither purely discrete nor purely continuous.

Normalising a Claim-Size Density A claim size X (in $000s) has density f(x) = c(10 - x) for 0 < x < 10. Find c, P(X > 4) and E[X].

Answer The density must integrate to 1:

\begin{align*}
1 &= \int_0^{10} c(10 - x)\,dx \\
  &= c\left[10x - \tfrac{x^2}{2}\right]_0^{10} \\
  &= 50c
\end{align*}

so c = 0.02. Then

\begin{align*}
P(X > 4) &= \int_4^{10} 0.02(10 - x)\,dx \\
         &= 0.02\left[10x - \tfrac{x^2}{2}\right]_4^{10} \\
         &= 0.02(50 - 32) \\
         &= 0.36
\end{align*}

\begin{align*}
E[X] &= \int_0^{10} 0.02\,x(10 - x)\,dx \\
     &= 0.02\left[5x^2 - \tfrac{x^3}{3}\right]_0^{10} \\
     &= 0.02(500 - 333.33) \\
     &= 3.33
\end{align*}

The mean claim is about $3,333 and 36% of claims exceed $4,000. P(X \ge 4) is the same 0.36 — the endpoint carries no probability.

A Deductible Creates a Point Mass A loss X is uniform on (0, 1{,}000). The policy pays Y = \max(X - 200,\ 0). Find P(Y = 0), P(Y \le 300) and E[Y], and decide whether Y is continuous.

Answer The policy pays nothing whenever the loss is below the deductible:

\begin{align*}
P(Y = 0) &= P(X \le 200) \\
         &= 0.2
\end{align*}

A single value carries probability 0.2, so Y is not continuous. It is mixed: a mass of 0.2 at zero plus a density of \tfrac{1}{1{,}000} on (0, 800).

\begin{align*}
P(Y \le 300) &= P(X \le 500) \\
             &= 0.5
\end{align*}

\begin{align*}
E[Y] &= \int_{200}^{1{,}000} (x - 200)\,\frac{1}{1{,}000}\,dx \\
     &= \frac{800^2/2}{1{,}000} \\
     &= 320
\end{align*}

The expected payment per loss is $320. Integrating a density alone would miss the 20% of probability sitting at zero — which is why exam questions ask for P(Y = 0) separately.