A continuous random variable x has a probability density function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\) Then the variance of x is:
Finding Variance of a Continuous Random Variable
To find the variance of a continuous random variable \(x\) with a given probability density function (PDF), \(f(x)\), we use the formula:
\(Var(x) = E[x^2] - (E[x])^2\)
where \(E[x]\) is the expected value (mean) of \(x\) and \(E[x^2]\) is the expected value of \(x^2\).
The given probability density function is:
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\)
Step-by-Step Calculation of Expected Value E[x]
The expected value \(E[x]\) for a continuous random variable is calculated using the integral:
\(E[x] = \int_{-\infty}^{\infty} x f(x) dx\)
Given the PDF is non-zero only for \(0 < x \le 1\), the integral becomes:
\(E[x] = \int_{0}^{1} x \cdot 6x(1 - x) dx\)
First, expand the expression inside the integral:
\(x \cdot 6x(1 - x) = 6x^2(1 - x) = 6x^2 - 6x^3\)
Now, integrate this expression from 0 to 1:
\(E[x] = \int_{0}^{1} (6x^2 - 6x^3) dx\)
Integrate term by term:
\(\int (6x^2 - 6x^3) dx = 6 \int x^2 dx - 6 \int x^3 dx = 6 \frac{x^3}{3} - 6 \frac{x^4}{4} = 2x^3 - \frac{3}{2}x^4\)
Evaluate the definite integral from 0 to 1:
\(E[x] = \left[ 2x^3 - \frac{3}{2}x^4 \right]_{0}^{1}\)
\(E[x] = \left( 2(1)^3 - \frac{3}{2}(1)^4 \right) - \left( 2(0)^3 - \frac{3}{2}(0)^4 \right)\)
\(E[x] = \left( 2 - \frac{3}{2} \right) - (0 - 0)\)
\(E[x] = \frac{4}{2} - \frac{3}{2} = \frac{1}{2}\)
So, the expected value \(E[x]\) is \(\frac{1}{2}\).
Step-by-Step Calculation of Expected Value E[x<sup>2</sup>]
The expected value \(E[x^2]\) for a continuous random variable is calculated using the integral:
\(E[x^2] = \int_{-\infty}^{\infty} x^2 f(x) dx\)
Given the PDF is non-zero only for \(0 < x \le 1\), the integral becomes:
\(E[x^2] = \int_{0}^{1} x^2 \cdot 6x(1 - x) dx\)
First, expand the expression inside the integral:
\(x^2 \cdot 6x(1 - x) = 6x^3(1 - x) = 6x^3 - 6x^4\)
Now, integrate this expression from 0 to 1:
\(E[x^2] = \int_{0}^{1} (6x^3 - 6x^4) dx\)
Integrate term by term:
\(\int (6x^3 - 6x^4) dx = 6 \int x^3 dx - 6 \int x^4 dx = 6 \frac{x^4}{4} - 6 \frac{x^5}{5} = \frac{3}{2}x^4 - \frac{6}{5}x^5\)
Evaluate the definite integral from 0 to 1:
\(E[x^2] = \left[ \frac{3}{2}x^4 - \frac{6}{5}x^5 \right]_{0}^{1}\)
\(E[x^2] = \left( \frac{3}{2}(1)^4 - \frac{6}{5}(1)^5 \right) - \left( \frac{3}{2}(0)^4 - \frac{6}{5}(0)^5 \right)\)
\(E[x^2] = \left( \frac{3}{2} - \frac{6}{5} \right) - (0 - 0)\)
To subtract the fractions, find a common denominator, which is 10:
\(E[x^2] = \frac{3 \cdot 5}{2 \cdot 5} - \frac{6 \cdot 2}{5 \cdot 2} = \frac{15}{10} - \frac{12}{10} = \frac{15 - 12}{10} = \frac{3}{10}\)
So, the expected value \(E[x^2]\) is \(\frac{3}{10}\).
Calculating the Variance Var(x)
Now we use the formula for variance:
\(Var(x) = E[x^2] - (E[x])^2\)
Substitute the calculated values for \(E[x]\) and \(E[x^2]\):
\(Var(x) = \frac{3}{10} - \left(\frac{1}{2}\right)^2\)
Calculate \(\left(\frac{1}{2}\right)^2\):
\(\left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}\)
Substitute this back into the variance formula:
\(Var(x) = \frac{3}{10} - \frac{1}{4}\)
To subtract these fractions, find a common denominator, which is 20:
\(Var(x) = \frac{3 \cdot 2}{10 \cdot 2} - \frac{1 \cdot 5}{4 \cdot 5}\)
\(Var(x) = \frac{6}{20} - \frac{5}{20}\)
\(Var(x) = \frac{6 - 5}{20} = \frac{1}{20}\)
The variance of the continuous random variable \(x\) is \(\frac{1}{20}\).
Let's compare this result with the given options:
- Option 1: \(\dfrac{1}{20}\)
- Option 2: \(\dfrac{1}{40}\)
- Option 3: \(\dfrac{1}{5}\)
- Option 4: \(\dfrac{1}{2}\)
Our calculated variance matches Option 1.
Summary of Variance Calculation Steps
| Step | Description | Formula / Calculation |
|---|---|---|
| 1 | Identify the PDF, \(f(x)\). | \(f(x) = 6x(1-x)\) for \(0 < x \le 1\) |
| 2 | Calculate the expected value \(E[x]\). | \(E[x] = \int_{0}^{1} x f(x) dx = \int_{0}^{1} (6x^2 - 6x^3) dx = \frac{1}{2}\) |
| 3 | Calculate the expected value \(E[x^2]\). | \(E[x^2] = \int_{0}^{1} x^2 f(x) dx = \int_{0}^{1} (6x^3 - 6x^4) dx = \frac{3}{10}\) |
| 4 | Calculate the variance \(Var(x)\). | \(Var(x) = E[x^2] - (E[x])^2 = \frac{3}{10} - \left(\frac{1}{2}\right)^2 = \frac{1}{20}\) |
Revision Table: Key Concepts in Variance and Expected Value
| Concept | Definition | Formula (Continuous RV) |
|---|---|---|
| Expected Value (Mean) | The average value of a random variable over many trials. | \(E[x] = \int_{-\infty}^{\infty} x f(x) dx\) |
| Expected Value of g(x) | The average value of a function of a random variable. | \(E[g(x)] = \int_{-\infty}^{\infty} g(x) f(x) dx\) |
| Variance | A measure of how spread out the values of a random variable are from its mean. | \(Var(x) = E[(x - E[x])^2] = E[x^2] - (E[x])^2\) |
| Probability Density Function (PDF) | A function describing the relative likelihood for a continuous random variable to take on a given value. The integral of the PDF over its support must equal 1. | \(\int_{-\infty}^{\infty} f(x) dx = 1\) |
Additional Information: Properties of Variance
- The variance of a constant \(c\) is zero: \(Var(c) = 0\).
- The variance of a linear transformation \(ax + b\) is: \(Var(ax + b) = a^2 Var(x)\). Note that adding a constant \(b\) does not change the variance.
- Variance is always non-negative: \(Var(x) \ge 0\).
- The square root of the variance is the standard deviation: \(SD(x) = \sqrt{Var(x)}\).
Important Questions from Mean and Variance of Random variables
X is a non-negative integer valued random variable with
\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)
Then, mean and variance of X are respectively
A random variable X has the distribution law as given below:
X
1
2
3
P(X = x)
0.3
0.4
0.3
The variance of the distribution is:
