Worksheet: Continuous Random Variables

In this worksheet, you will find questions and exercises on continuous random variables and common continuous distributions. After the first question in specific distribution topics, a relevant calculator from my site will be displayed.

General Continuous Distributions: Concepts

1. What are the two main properties that a function \(f(x)\) must satisfy to be a valid probability density function (PDF) for a continuous random variable \(X\)?

A function \(f(x)\) is a valid PDF if:

\(f(x) \ge 0\) for all \(x\) (the function is non-negative).
\(\int_{-\infty}^{\infty} f(x) \, dx = 1\) (the total area under the curve is 1).

2. If \(X\) is a continuous random variable with PDF \(f(t)\), how is its cumulative distribution function (CDF), \(F(x)\), defined? What does \(F(x)\) represent?

The CDF, \(F(x)\), is defined as:

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

\(F(x)\) represents the probability that the random variable \(X\) takes on a value less than or equal to \(x\).

3. How do you calculate the probability \(P(a < X \le b)\) for a continuous random variable \(X\) using its PDF \(f(x)\) and its CDF \(F(x)\)? What is \(P(X=c)\) for any constant \(c\)?

Using the PDF:

\[P(a < X \le b) = \int_{a}^{b} f(x) \, dx\]

Using the CDF:

\[P(a < X \le b) = F(b) - F(a)\]

For any continuous random variable \(X\) and any constant \(c\), the probability \(P(X=c) = 0\). This means that for continuous variables, \(P(a < X \le b) = P(a \le X \le b) = P(a < X < b) = P(a \le X < b)\).

4. How is the expected value (or mean) \(E[X]\) of a continuous random variable \(X\) with PDF \(f(x)\) calculated?

The expected value \(E[X]\), denoted by \(\mu\), is calculated as:

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

5. Define the variance \(\text{Var}(X)\) of a continuous random variable \(X\) with PDF \(f(x)\) and mean \(\mu\). How can it be calculated using expected values? What is the standard deviation?

The variance \(\text{Var}(X)\), denoted by \(\sigma^2\), is defined as:

\[\text{Var}(X) = \sigma^2 = E[(X-\mu)^2] = \int_{-\infty}^{\infty} (x-\mu)^2 f(x) \, dx\]

It can also be calculated using the formula:

\[\text{Var}(X) = E[X^2] - (E[X])^2 = E[X^2] - \mu^2\]

where \(E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx\).
The standard deviation, \(\sigma\), is the square root of the variance: \(\sigma = \sqrt{\text{Var}(X)}\).

Exponential Distribution

1. A random variable \(X\) follows an exponential distribution with rate parameter \(\lambda > 0\). What is its PDF? What is a key property of this distribution often referred to as?

The Probability Density Function (PDF) of an exponential distribution is:

\[ f(x; \lambda) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0 \\ 0 & x < 0 \end{cases} \]

A key property of the exponential distribution is **memorylessness**. This means that for \(s, t \ge 0\):

\[P(X > s+t | X > s) = P(X > t)\]

In words, the probability that an item lasts an additional \(t\) units of time, given it has already lasted \(s\) units, is the same as the probability that a new item lasts \(t\) units of time.

2. The lifetime of a certain type of electronic component is exponentially distributed with a mean lifetime of 500 hours. What is the rate parameter \(\lambda\)? What is the probability that the component lasts for at least 700 hours?

For an exponential distribution, the mean (expected value) is \(E[X] = 1/\lambda\).
Given mean = 500 hours, we have \(500 = 1/\lambda\), so the rate parameter \(\lambda = 1/500 = 0.002\).
The Cumulative Distribution Function (CDF) is \(F(x) = P(X \le x) = 1 - e^{-\lambda x}\) for \(x \ge 0\).
We want to find the probability that the component lasts for at least 700 hours, which is \(P(X \ge 700)\).
\(P(X \ge 700) = 1 - P(X < 700)\). Since it's a continuous distribution, \(P(X < 700) = P(X \le 700) = F(700)\).
So, \(P(X \ge 700) = 1 - F(700) = 1 - (1 - e^{-\lambda \times 700}) = e^{-0.002 \times 700} = e^{-1.4}\).
\(e^{-1.4} \approx 0.2466\).
Thus, the probability that the component lasts for at least 700 hours is approximately 0.2466.

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Continuous Uniform Distribution

1. A random variable \(X\) is uniformly distributed on the interval \([a, b]\). What is its Probability Density Function (PDF)?

The PDF of a continuous uniform distribution on the interval \([a, b]\) is:

\[ f(x; a, b) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases} \]

2. The time (in minutes) a shuttle bus takes to arrive at a particular stop is uniformly distributed between 0 and 15 minutes. What is the probability that a person arriving at the stop will have to wait more than 10 minutes for the bus?

Let \(X\) be the arrival time in minutes. \(X \sim U(0, 15)\). So, \(a=0\) and \(b=15\).
The PDF is \(f(x) = \frac{1}{15-0} = \frac{1}{15}\) for \(0 \le x \le 15\).
We want to find the probability \(P(X > 10)\). Since the bus arrives at time \(X\), waiting more than 10 minutes means the bus arrives between 10 and 15 minutes.

\[P(X > 10) = \int_{10}^{15} \frac{1}{15} \, dx = \frac{1}{15} [x]_{10}^{15} = \frac{1}{15} (15 - 10) = \frac{5}{15} = \frac{1}{3}\]

Alternatively, using the CDF \(F(x) = \frac{x-a}{b-a}\) for \(a \le x \le b\).
\(P(X > 10) = 1 - P(X \le 10) = 1 - F(10) = 1 - \frac{10-0}{15-0} = 1 - \frac{10}{15} = 1 - \frac{2}{3} = \frac{1}{3}\).

3. For a continuous uniform random variable \(X \sim U(a,b)\), what are its expected value \(E[X]\) and variance \(\text{Var}(X)\)?

For \(X \sim U(a,b)\):
Expected Value:

\[E[X] = \frac{a+b}{2}\]

Variance:

\[\text{Var}(X) = \frac{(b-a)^2}{12}\]

General Continuous Distributions: Concepts

Exponential Distribution

Continuous Uniform Distribution

Short Quiz: Continuous Distributions