Annual losses on an insurance policy have a distribution with the probability density function 
f(x)={(e^((-x^2)/200))/(5(√2π)),&x>0@0,&otherwise)┤ 
Calculate the expected value of X.

The probability density function (PDF) of annual losses on an insurance policy is given by:

\[
f(x) = \frac{e^{-x^2 / 200}}{5 \sqrt{2\pi}}, \quad x > 0
\]

And \( f(x) = 0 \) for \( x \leq 0 \).

To calculate the expected value \( E(X) \) of \( X \), we use the formula for the expected value of a continuous random variable:

\[
E(X) = \int_0^{\infty} x f(x) \, dx
\]

Substituting the given PDF \( f(x) \):

\[
E(X) = \int_0^{\infty} x \cdot \frac{e^{-x^2 / 200}}{5 \sqrt{2\pi}} \, dx
\]

We can factor out the constant:

\[
E(X) = \frac{1}{5 \sqrt{2\pi}} \int_0^{\infty} x e^{-x^2 / 200} \, dx
\]

Now, we recognize that the integral has the form of a standard Gaussian integral. To simplify the computation, we use the following substitution:

Let \( u = \frac{x^2}{200} \), so \( du = \frac{x dx}{100} \), or \( x dx = 100 du \).

Substituting into the integral:

\[
E(X) = \frac{1}{5 \sqrt{2\pi}} \cdot 100 \int_0^{\infty} e^{-u} \, du
\]

The integral of \( e^{-u} \) from 0 to infinity is 1. Therefore, we have:

\[
E(X) = \frac{100}{5 \sqrt{2\pi}} \cdot 1 = \frac{100}{5 \sqrt{2\pi}} = \frac{20}{\sqrt{2\pi}}
\]

Since \( \sqrt{2\pi} \approx 2.5066 \), the expected value is approximately:

\[
E(X) \approx \frac{20}{2.5066} \approx 7.98
\]

Thus, the expected value of \( X \) is approximately \( 7.98 \).

Would you like more details or have any questions?

Here are five related questions to explore this concept further:
1. How would the expected value change if the variance in the distribution increased?
2. Can we calculate the variance of \( X \) from this PDF?
3. How does this result compare with the expected value of a normal distribution?
4. What other types of distributions could be used to model insurance losses?
5. How do tail probabilities affect the expected value of insurance losses?

**Tip:** Substitutions can simplify integrals, especially when Gaussian-like terms appear in the function.

Annual losses on an insurance policy have a distribution with the probability density function f(x) = (e^((-x^2)/200))/(5√(2π)), x > 0. Calculate the expected value of X.

Explore how to calculate the expected value of annual insurance losses given a Gaussian-like probability density function. Learn about continuous random variables, Gaussian integrals, and the detailed steps to solve for E(X) with an example PDF. Ideal for university-level probability students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation