Problem 5 - Calculus and its applications

Question

For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{2}{3} x, \quad[1,2] $$

Step-by-Step Solution

Answer

\( \mathrm{E}(x) = \frac{7}{9}, \mathrm{E}(x^2) = \frac{5}{2}, \) mean is \( \frac{7}{9}, \) variance is approximately 1.617, and standard deviation is approximately 1.271.

Step 1: Understanding the Function

We are given a probability density function (pdf) \( f(x) = \frac{2}{3}x \), which is defined over the interval \([1,2]\). We need to use this function to find several statistical measures: expectation of \( x \) and \( x^2 \), mean, variance, and standard deviation.

Step 2: Finding Expectation \( \mathrm{E}(x) \)

The expectation, \( \mathrm{E}(x) \), is calculated as \( \int_{a}^{b} x f(x) \, dx \). Here it becomes: \[ \mathrm{E}(x) = \int_{1}^{2} x \left(\frac{2}{3} x\right) \, dx = \int_{1}^{2} \frac{2}{3} x^2 \, dx. \] Calculate the integral: \[ \int \frac{2}{3} x^2 \, dx = \frac{2}{3} \times \frac{x^3}{3} = \frac{x^3}{9}. \] Evaluate this from 1 to 2: \[ \left(\frac{2^3}{9}\right) - \left(\frac{1^3}{9}\right) = \frac{8}{9} - \frac{1}{9} = \frac{7}{9}. \] Therefore, \( \mathrm{E}(x) = \frac{7}{9}. \)

Step 3: Finding Expectation \( \mathrm{E}(x^2) \)

The expectation of \( x^2 \), \( \mathrm{E}(x^2) \), is found by evaluating \( \int_{1}^{2} x^2 f(x) \, dx \): \[ \mathrm{E}(x^2) = \int_{1}^{2} x^2 \left(\frac{2}{3} x\right) \, dx = \int_{1}^{2} \frac{2}{3} x^3 \, dx. \] Calculate the integral: \[ \int \frac{2}{3} x^3 \, dx = \frac{2}{3} \times \frac{x^4}{4} = \frac{x^4}{6}. \] Evaluate it from 1 to 2: \[ \left(\frac{2^4}{6}\right) - \left(\frac{1^4}{6}\right) = \frac{16}{6} - \frac{1}{6} = \frac{15}{6} = \frac{5}{2}. \] Thus, \( \mathrm{E}(x^2) = \frac{5}{2}. \)

Step 4: Calculating the Mean

The mean of the distribution is simply \( \mathrm{E}(x) \), which we have calculated as \( \frac{7}{9}. \)

Step 5: Calculating the Variance

The variance is given by \( \mathrm{Var}(x) = \mathrm{E}(x^2) - [\mathrm{E}(x)]^2 \). Plug in the values: \[ \mathrm{Var}(x) = \frac{5}{2} - \left(\frac{7}{9}\right)^2 = \frac{5}{2} - \frac{49}{81}. \] First express \( \frac{5}{2} \) in terms of 81: \[ \frac{5}{2} = \frac{202.5}{81} = \frac{405}{162}. \] Therefore, \[ \mathrm{Var}(x) = \frac{405}{162} - \frac{49}{81} = \frac{405}{162} - \frac{98}{162} = \frac{307}{162} = \frac{307}{162}. \] Thus, \( \mathrm{Var}(x) \) approximately equals 1.617.

Step 6: Finding the Standard Deviation

The standard deviation is the square root of the variance. So, calculate as: \[ \sigma = \sqrt{\frac{307}{162}}. \] This is approximately 1.271.

Key Concepts

Expectation

The concept of expectation is fundamental in probability and statistics. It represents the average or expected value of a random variable. For continuous distributions, the expectation, denoted as \( \mathrm{E}(x) \), is calculated using an integral over the probability density function (pdf) times the variable itself. In our example, the pdf is \( f(x) = \frac{2}{3}x \) over the interval \([1,2]\). To find \( \mathrm{E}(x) \), we calculate:
\[ \mathrm{E}(x) = \int_{1}^{2} x \, f(x) \, dx = \int_{1}^{2} \frac{2}{3} x^2 \, dx = \frac{7}{9}. \]
The process involves integrating the function \( x \, f(x) \) from 1 to 2, which gives us the expected value or average if we were to observe many values of \( x \) from the distribution.

Mean

The mean, frequently interchangeable with expectation, refers to the central tendency or the average of a probability distribution.
For a continuous random variable, it's calculated as the expected value of \( x \). In our case, since we have already calculated \( \mathrm{E}(x) = \frac{7}{9} \), this value is the mean of our distribution.
The mean provides a measure of where the values of the random variable tend to cluster. It's important to note that for symmetrical distributions, the mean is located at the center, but this need not be the case for all distributions.

Variance

Variance measures the spread or dispersion of a set of values around the mean. It shows how much the values of a random variable are likely to differ from the mean.
To calculate the variance \( \mathrm{Var}(x) \), we use the formula:

• \( \mathrm{Var}(x) = \mathrm{E}(x^2) - [\mathrm{E}(x)]^2 \).

From our calculations:

• \( \mathrm{E}(x^2) = \frac{5}{2} \)
• \( \mathrm{E}(x) = \frac{7}{9} \)

Substituting these into the variance formula gives:

• \( \frac{5}{2} - \left( \frac{7}{9} \right)^2 = \frac{307}{162} \)

This reflects how the data points spread around the mean \( \frac{7}{9} \). A higher variance indicates more spread out data values, while a lower variance signals that they are closely clustered.

Standard Deviation

The standard deviation is the square root of the variance and provides a measure of the amount of variation or dispersion in a set of values.
Since variance in our example is \( \mathrm{Var}(x) = \frac{307}{162} \), the standard deviation \( \sigma \) becomes:
\[ \sigma = \sqrt{\frac{307}{162}} \]
This value, approximately 1.271, is more intuitive to interpret compared to variance because it's expressed in the same unit as the data points and their mean.
In simple terms, the standard deviation tells us that, on average, the data points differ from the mean by about 1.271 units in our given distribution.