Suppose that X is a continuous random variable with probability density function given by: f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\) Find the mean of X.
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Mean Calculation for Continuous Random Variable X
To determine the mean of X, also known as its expected value, for a continuous random variable, we must use its probability density function (PDF). The question provides a piecewise probability density function for X and asks us to find its mean.
Expected Value Formula
For any continuous random variable X with a probability density function \(f(x)\), the mean or expected value, denoted as \(E[X]\), is calculated by integrating \(x \cdot f(x)\) over the entire range of X where \(f(x)\) is non-zero. The formula is:
\[ E[X] = \int_{-\infty}^{\infty} x \cdot f(x) \, dx \]
In this particular problem, the probability density function \(f(x)\) is given by:
- \(f(x) = \frac{x}{8}\) for \(x \in \left[ {0,2} \right)\)
- \(f(x) = \frac{1}{4}\) for \(x \in \left[ {2,4} \right)\)
- \(f(x) = -\frac{x}{8} + \frac{3}{4}\) for \(x \in \left[ {4,6} \right)\)
- \(f(x) = 0\) for all other values of \(x\).
Integral Setup for Mean of X
Since the probability density function \(f(x)\) is defined differently over various intervals, we will set up the integral for \(E[X]\) by summing the integrals over each specific interval where \(f(x)\) is defined and non-zero:
\[ E[X] = \int_{0}^{2} x \left( \frac{x}{8} \right) dx + \int_{2}^{4} x \left( \frac{1}{4} \right) dx + \int_{4}^{6} x \left( -\frac{x}{8} + \frac{3}{4} \right) dx \]
Let's evaluate each of these three integrals step by step.
Integral Terms Calculation
First Integral: Interval \(\left[ {0,2} \right)\)
For the first interval, \(f(x) = \frac{x}{8}\):
\[ \text{Integral}_1 = \int_{0}^{2} x \left( \frac{x}{8} \right) dx = \int_{0}^{2} \frac{x^2}{8} dx \]
Perform the integration:
\[ = \frac{1}{8} \left[ \frac{x^3}{3} \right]_{0}^{2} \]
Apply the limits of integration:
\[ = \frac{1}{8} \left( \frac{2^3}{3} - \frac{0^3}{3} \right) = \frac{1}{8} \left( \frac{8}{3} - 0 \right) = \frac{1}{8} \cdot \frac{8}{3} = \frac{1}{3} \]
Second Integral: Interval \(\left[ {2,4} \right)\)
For the second interval, \(f(x) = \frac{1}{4}\):
\[ \text{Integral}_2 = \int_{2}^{4} x \left( \frac{1}{4} \right) dx = \int_{2}^{4} \frac{x}{4} dx \]
Perform the integration:
\[ = \frac{1}{4} \left[ \frac{x^2}{2} \right]_{2}^{4} \]
Apply the limits of integration:
\[ = \frac{1}{4} \left( \frac{4^2}{2} - \frac{2^2}{2} \right) = \frac{1}{4} \left( \frac{16}{2} - \frac{4}{2} \right) = \frac{1}{4} (8 - 2) = \frac{1}{4} (6) = \frac{6}{4} = \frac{3}{2} \]
Third Integral: Interval \(\left[ {4,6} \right)\)
For the third interval, \(f(x) = -\frac{x}{8} + \frac{3}{4}\):
\[ \text{Integral}_3 = \int_{4}^{6} x \left( -\frac{x}{8} + \frac{3}{4} \right) dx = \int_{4}^{6} \left( -\frac{x^2}{8} + \frac{3x}{4} \right) dx \]
Perform the integration:
\[ = \left[ -\frac{x^3}{8 \cdot 3} + \frac{3x^2}{4 \cdot 2} \right]_{4}^{6} = \left[ -\frac{x^3}{24} + \frac{3x^2}{8} \right]_{4}^{6} \]
Apply the limits of integration:
\[ = \left( -\frac{6^3}{24} + \frac{3 \cdot 6^2}{8} \right) - \left( -\frac{4^3}{24} + \frac{3 \cdot 4^2}{8} \right) \]
\[ = \left( -\frac{216}{24} + \frac{3 \cdot 36}{8} \right) - \left( -\frac{64}{24} + \frac{3 \cdot 16}{8} \right) \]
Simplify the terms:
\[ = \left( -9 + \frac{108}{8} \right) - \left( -\frac{8}{3} + \frac{48}{8} \right) \]
\[ = \left( -9 + \frac{27}{2} \right) - \left( -\frac{8}{3} + 6 \right) \]
Combine terms within each parenthesis:
\[ = \left( -\frac{18}{2} + \frac{27}{2} \right) - \left( -\frac{8}{3} + \frac{18}{3} \right) \]
\[ = \frac{9}{2} - \frac{10}{3} \]
Find a common denominator (6) and perform the subtraction:
\[ = \frac{9 \cdot 3}{2 \cdot 3} - \frac{10 \cdot 2}{3 \cdot 2} = \frac{27}{6} - \frac{20}{6} = \frac{7}{6} \]
Mean Summation Calculation
Finally, we sum the results of all three integrals to find the total mean of X:
\[ E[X] = \text{Integral}_1 + \text{Integral}_2 + \text{Integral}_3 \]
\[ E[X] = \frac{1}{3} + \frac{3}{2} + \frac{7}{6} \]
To add these fractions, we find a common denominator, which is 6:
\[ E[X] = \frac{1 \cdot 2}{3 \cdot 2} + \frac{3 \cdot 3}{2 \cdot 3} + \frac{7}{6} \]
\[ E[X] = \frac{2}{6} + \frac{9}{6} + \frac{7}{6} \]
\[ E[X] = \frac{2 + 9 + 7}{6} = \frac{18}{6} = 3 \]
Mean Conclusion
The mean of X, or its expected value, is 3.
Here's a summary of the individual integral calculations:
| Interval | Function \(f(x)\) | Integral \(\int x f(x) dx\) | Calculated Value |
|---|---|---|---|
| \(\left[ {0,2} \right)\) | \(\frac{x}{8}\) | \(\int_{0}^{2} \frac{x^2}{8} dx\) | \(\frac{1}{3}\) |
| \(\left[ {2,4} \right)\) | \(\frac{1}{4}\) | \(\int_{2}^{4} \frac{x}{4} dx\) | \(\frac{3}{2}\) |
| \(\left[ {4,6} \right)\) | \(-\frac{x}{8} + \frac{3}{4}\) | \(\int_{4}^{6} \left(-\frac{x^2}{8} + \frac{3x}{4}\right) dx\) | \(\frac{7}{6}\) |
Adding the calculated values gives the final mean of X:
\[ E[X] = \frac{1}{3} + \frac{3}{2} + \frac{7}{6} = \frac{2}{6} + \frac{9}{6} + \frac{7}{6} = \frac{18}{6} = 3 \]
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
- The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is
