Last updated on 8 September 2025
A continuous random variable is a random variable that can have a range of values within a specific range. It is used to represent measurements like weight, height, and time. In this article, we will explore the concepts and properties of continuous random variables in detail.
A continuous random variable has an infinite number of possible values within a certain range. In statistics and probability theory, a continuous random variable plays a crucial role where the results are measured instead of being counted. A random variable is considered continuous if it can take any value within a specific interval. It is characterized by its possible values, which are infinite or uncountable. Also, using a function known as the probability density function, we can determine the probability that its value will fall inside a specific interval.
To understand the probability distributions and conduct statistical analysis, we need to comprehend the various properties of a continuous random variable. Several important properties distinguish the continuous random variable from the discrete random variable. They include:
A probability density function f(x) defines a continuous random variable x. The relative probability for x to be a certain value x is represented by the PDF f(x). The PDF needs to meet two conditions. The first one is f(x) ≥ 0 for all x. It means the probability density function (PDF) must always be non-negative but can be zero for some values. The next condition is that the total probability over all possible values of x must be 1, meaning the sum of all probability values must be 100% or 1. This is written as:
\(\int_{-\infty}^{\infty} f(x)\, dx = 1 \)
The cumulative distribution function (CDF), represented as F(x), shows the likelihood that a continuous random variable X will take a value that is less than or equal to x:
\(F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt \)
The CDF is non-decreasing and continuous. The CDF approaches 0 as x moves closer to negative infinity.
lim F(x) = 0
x→∞
Likewise, the CDF approaches 1 as x moves closer to positive infinity.
lim F(x) = 1
x→∞
The moment generating function (MGF) of a continuous random variable X is represented as Mx (t). It is defined as:
\(M_X(t) = E\!\left(e^{tX}\right) = \int_{-\infty}^{\infty} e^{tx} f(x)\,dx \)
The MGF can be used to find all the moments of X, including its mean and variance if it exists.
The characteristic function of a continuous random variable X is expressed as ϕX(t). It is the Fourier transform of the probability density function (PDF).
\(\varphi_X(t) = E\!\left(e^{itX}\right) = \int_{-\infty}^{\infty} e^{itx} f(x)\,dx\)
It determines the distribution of X uniquely.
A continuous random variable X with PDF f(x) has the following expectation or mean:
\(E(X) = \int_{-\infty}^{\infty} x f(x)\,dx \)
The formula represents the expected value of X.
The variance of X measures the average of the squared deviations of the random variable from the mean. It is defined as:
\(\operatorname{Var}(X) = E\!\left[(X - E(X))^{2}\right] = \int_{-\infty}^{\infty} (x - \mu)^{2} f(x)\,dx \)
Here, μ = E (X) is the mean.
Where X's second moment is represented by E(X2) and it is denoted as:
\(E(X^2) = \int_{-\infty}^{\infty} x^2 f(x)\,dx \)
The probabilities associated with a continuous random variable are described by the probability density function (PDF) and cumulative probability function (CDF). Here are some key formulas related to continuous random variables.
Here, F(x) is the cumulative distribution function.
A random experiment’s numerical outcome is called a random variable. The two types of random variables are discrete random variable and continuous random variable. Let us look at the differences between them:
Continuous Random Variable
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Discrete Random Variable
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Can have any value within a specified range
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Can only have particular and separate values
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The possible values are infinite within a certain range. Example: all real values between 1 and 2
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The possible values are finite or countably infinite. Example: 1, 2, 3, and so on.
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A probability density function (PDF) describes a continuous random variable
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A probability mass function describes a discrete random variable
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The probability of a single value is zero (P (X = x) = 0
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The probability of a single value is non-zero (P (X = x) > 0
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It is represented by a smooth curve
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It is represented by bar graphs
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Examples include time, distance, temperature, height, and weight
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Examples include the number of children, the number of flowers, and the results of a die roll
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We use continuous random variables to model situations that involve measurements. For example, if we want to know the possible amount of rainfall over a year or temperature on any given day. Following are the significant continuous random variables linked to certain probability distributions:
Uniform Random Variable
A uniform random variable represents a uniform distribution, which describes events with equal chances of happening. A uniform random variable’s PDF is as follows:
Otherwise, a ≤ x ≤ b
Here, a and b are the lower and upper bounds of the distribution.
Normal Random Variable
A normal random variable is a continuous random variable that is used to model a normal distribution. If a normal distribution’s parameters are expressed as X ~ N (μ, σ2), then the following is the formula for the PDF:
Here, μ is the mean
σ is the standard deviation
σ2 is the variance
Exponential Random Variable
An exponential distribution is a continuous probability distribution used to model processes in which a specific number of events occur continuously and independently at a constant average rate λ, where λ ≥ 0. The exponential random variable follows an exponential distribution. The following is the PDF of an exponential random variable:
Within a specified range, a continuous random variable can have an infinite number of values. This concept is widely used in situations where measurements are involved. Here are some real-world significance of continuous random variables.
A continuous random variable is a random variable that can take any value within a specific range. Knowing the key properties and concepts of this variable helps us to solve various mathematical problems, and it improves our problem-solving skills. Here are some of the common mistakes and their helpful solutions to avoid these common errors.
A random variable X has the probability density function (PDF): f (x) = k(5−x), 0 ≤ x ≤5. Find the value of k.
0.08
A function must adhere to the following basic rule in order to be considered a valid probability density function (PDF):
\(\int_{-\infty}^{\infty} f(x)\,dx = 1 \)
As we know, the given PDF is defined for 0 ≤ x ≤5, then the integral is taken over this range:
\(\int_{0}^{5} f(x)\,dx = 1 \)
Next, we can substitute f(x) = k (5 - x):
So, now we have to factor out the constant k:
\(k \int_{0}^{5} (5 - x)\,dx = 1 \)
Then, we have to integrate (5 - x) term by term:
∫ (5 - x) dx = ∫ 5 dx - ∫ x dx
Here, we have to use the basic integration rules:
∫ 5 dx = 5x
∫ x dx = x2 / 2
So the antiderivative is:
\(\int (5 - x)\,dx = 5x - \frac{x^2}{2} \)
Next, we evaluate the definite integral from 0 to 5:
\(\begin{align*} \int_{0}^{5} (5 - x)\,dx &= \left[ 5x - \frac{x^2}{2} \right]_0^5 \\ &= \left( 5(5) - \frac{5^2}{2} \right) - \left( 5(0) - \frac{0^2}{2} \right) \\ &= (25 - 12.5) - 0 \\ &= 12.5 \end{align*} \)
Substitute back to solve for k:
So, k (12.5) = 1
k = 1 / 12.5
k = 0.08
Hence, the value of k is 0.08.
For a continuous random variable with PDF: f (x)=4x2, 0 ≤ x ≤ 1. Find the value of E (X).
1
Given: f (x) = 4x2, 0 \(\le \) x \(\le \) 1
We have to find E (X) of X. Here, we can use the formula:
\(E(X) = \int_{a}^{b} x f(x)\,dx\)
Where a = 0 and b = 1
Now, we have to substitute f (x)
\(E(X) = \int_{0}^{1} x (4x^2)\,dx = \int_{0}^{1} 4x^3\,dx\)
Integrating,
\(\int 4x^3\,dx = x^4 + C\)
Evaluating the definite integral from 0 to 1,
\(E(X) = \left[ x^4 \right]_0^1 = 1^4 - 0^4 = 1 \)= 1
Find the mean E(X) for the uniform distribution U (2,10).
6
The formula for the mean of a uniform distribution U (a, b) is:
E (X) = (a + b) / 2
From U (2,10), we have a = 2 and b = 10, so:
E (X) = (2 + 10) / 2
= 12 / 2 = 6
Hence, the mean is 6.
Find the median m such that P (X ≤ m) = 0.6 for the uniform distribution U (0, 5).
3
Here, to find the value of the median, we have to use the formula:
F(m) = (m - a) / (b - a)
From U (0,5), this becomes:
F(m) = (m - 0) / 5 = m / 5
Setting F(m) = 0.6
m / 5 = 0.6
Next, we can solve for m:
m = 0.6 × 5
m = 3
The median is 3.
Find the median m such that P (X ≤ m) = 0.5 for the uniform distribution U (2, 8).
5
F(m) = (m - a) / (b - a)
From U (2,8), this becomes:
F(m) = (m - 2) / (8 - 2) = (m - 2) / 6
Setting F(m) = 0.5:
(m - 2) / 6 = 0.5
Next, we can solve for m:
m - 2 = 0.5 × 6
m - 2 = 3
m = 5
So, the median is 5.
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!