Question

A nonnegative function \(f\) is called a probability density function if \(\int_{-\infty}^{\infty} f(t) d t=1 .\) The probability that \(x\) lies between \(a\) and \(b\) is given by \(P(a \leq x \leq b)=\int_{a}^{b} f(t) d t\) The expected value of \(x\) is given by \(E(x)=\int_{-\infty}^{\infty} t f(t) d t\). Show that the nonnegative function is a probability density function, (b) find \(P(0 \leq x \leq 4),\) and (c) find \(E(x)\).$$ f(t)=\left\\{\begin{array}{ll} \frac{2}{5} e^{-2 t / 5}, & t \geq 0 \\ 0, & t<0 \end{array}\right. $$

Short answer

Therefore, \( f(t) \) is a probability density function. The probability that \( x \) lies between 0 and 4 is approximately 0.8193, and the expected value of \( x \) equals to 2.5.

Step-by-step solution

Verification of a Probability Density Function

To show that a given function is a probability density function, it must satisfy the condition that the integral of it from negative infinity to infinity equals to 1. Apply given \( f(t) \) on that, \( \int_{-\infty}^{\infty} f(t) dt = \int_{-\infty}^{0} 0 dt + \int_{0}^{\infty} \frac{2}{5} e^{-2t/5} dt = [0 - 0] + (-e^{-2t/5})|_{0}^{\infty} = 1 \)

Calculate the Probability for 0

Next, the probability that \( x \) lies between 0 and 4 needs to be calculated, which can be obtained through \( \int_{0}^{4} f(t) dt = \int_{0}^{4} \frac{2}{5} e^{-2t/5} dt = (-e^{-2t/5})|_{0}^{4} = 1 - e^{-8/5} \approx 0.8193 \)

Calculate the Expected Value of x

The expected value of \( x \) can be calculated by \( E(x) = \int_{-\infty}^{\infty} t f(t) dt = \int_{-\infty}^{0} 0 dt + \int_{0}^{\infty} t * \frac{2}{5} e^{-2t/5} dt = [0 - 0] + \frac{5}{2} = 2.5 \)

Key concepts

Expected Value

The expected value, often represented by \( E(x) \), is a fundamental concept in probability and statistics, especially when it comes to continuous random variables. It conveys the long-term average outcome of a random variable after many trials of a random process.

For a continuous random variable with a probability density function (PDF) \( f(t) \), the expected value is calculated using the integral \( E(x) = \int_{-\infty}^{\infty} t f(t) dt \), which represents the weighted average of all possible values. Here, each value \( t \) is weighted by its probability density \( f(t) \).

When computing the expected value, one typically performs a definite integral over the entire range of the random variable, considering the function's non-negativity and the area under the curve being equal to 1 to satisfy the properties of a PDF.

Definite Integrals

Definite integrals are a cornerstone in calculus, providing the net area under a curve within a specified interval \( [a, b] \). In the context of probability, the definite integral of a PDF over an interval \( [a, b] \) gives the probability that a random variable \( x \) falls within that range, formally expressed as \( P(a \leq x \leq b) = \int_{a}^{b} f(t) dt \).

To ensure a function is a PDF, one must integrate it over its entire domain and confirm the result equals 1, which establishes that it encompasses all the probabilities for the random variable. This process may involve integrating separate pieces of the function over corresponding intervals, especially if the function has different expressions over different ranges, as shown in the step-by-step solution provided.

Exponential Functions

Exponential functions play a vital role in many fields, including probability where they often describe the time until an event occurs, such as radioactive decay or life expectancy. In probability density functions, they are characterized by their rapidly decreasing nature, symbolized by \( f(t) = ae^{-bt} \) for some constants \( a \) and \( b \).

The base of the natural exponential function is Euler's number \( e \), approximately equal to 2.71828, which serves as the foundation of natural logarithms. The importance of these functions in probability comes from their 'memoryless' property, where the likelihood of an event occurring in the next interval is unaffected by how much time has already passed.