Problem 107 - An Introduction to Mathematical Statistics and Its Applications

Question

Find the variance of \(Y\) if $$ f_{Y}(y)= \begin{cases}\frac{3}{4}, & 0 \leq y \leq 1 \\ \frac{1}{4}, & 2 \leq y \leq 3 \\ 0, & \text { elsewhere }\end{cases} $$

Step-by-Step Solution

Answer

The variance of the random variable \(Y\) is \(\frac{1}{16}\)

Step 1: Calculation of Expectation

First, find expectation \(E[Y]\). The expectation is computed as the weighted average of the random variable. In this case \(E[Y]\) would computed by integrating product of \(y\) and its pdf over the range of the random variable: \[E[Y] = \int_0^1 y*f_Y(y) \, dy + \int_2^3 y*f_Y(y) \, dy\]Substituting and calculating we get:\[E[Y] = \int_0^1 y*(\frac{3}{4}) \, dy + \int_2^3 y*(\frac{1}{4}) \, dy\] \[E[Y] = \frac{1}{2} + \frac{5}{4} = \frac{7}{4}\]

Step 2: Calculation of Expectation of square

Next, find the expectation of square \(E[Y^2]\). It is computed by integrating the product of square of \(y\) and its probability function over the entire range:\[E[Y^2] = \int_0^1 y^2*f_Y(y) \, dy + \int_2^3 y^2*f_Y(y) \, dy\]Substituting and calculating we get:\[E[Y^2] = \int_0^1 y^2*(\frac{3}{4}) \, dy + \int_2^3 y^2*(\frac{1}{4}) \, dy\]\[E[Y^2] = \frac{1}{4} + \frac{9}{4} = \frac{5}{2}\]

Step 3: Calculation of Variance

Finally, calculate the variance of \(Y\). Variance is the expectation of the squared deviation of a random variable from its mean. It is computed using the formula \(Var[Y] = E[Y^2] - (E[Y])^2\). Substituting the values we found we get:\[Var[Y] = \frac{5}{2} - (\frac{7}{4})^2\]\[Var[Y] = \frac{5}{2} - \frac{49}{16} = \frac{1}{16}\]

Key Concepts

Variance

Variance is a measure of how much a set of values is spread out. In simple terms, it tells us how much the values of a random variable differ from the average value, or expectation. To calculate variance, we need to know two things:

• The mean or expectation of the random variable
• The expectation of the squared values of the random variable

The formula for variance is given by:\[Var(Y) = E[Y^2] - (E[Y])^2\]where:

• \(E[Y]\) is the expected value of the random variable \(Y\)
• \(E[Y^2]\) is the expected value of the square of the random variable \(Y\)

By using this formula, we can gauge the degree to which \(Y\) deviates from its mean value.

Expectation

Expectation, or the expected value, is a key concept in statistics that provides a measure of the center or average of the distribution of a random variable. In a sense, it tells us where the values of the random variable tend to cluster. For a continuous random variable like in our example, the expectation \(E[Y]\) is calculated by integrating the product of the random variable \(y\) and its probability density function (pdf) over all possible values of \(y\). It can be represented mathematically as:\[E[Y] = \int_{a}^{b} y \cdot f_Y(y) \, dy\]where \(a\) and \(b\) are the limits of the random variable. The expectation provides insight into the average outcome one can anticipate over repeated trials or observations.

Probability Density Function

The Probability Density Function (pdf) is a fundamental tool in probability and statistics. It describes the likelihood of a random variable to take on a particular value. Unlike discrete probabilities, which sum up to 1, a pdf integrates to 1 over the range of the random variable.The pdf is used to determine probabilities over an interval of the random variable. For a continuous random variable \(Y\) and pdf \(f_Y(y)\), the probability that \(Y\) falls within an interval \([a, b]\) is quantified by:\[P(a \leq Y \leq b) = \int_{a}^{b} f_Y(y) \, dy\]The pdf is critical for performing calculations of expectation and variance, as it defines the distribution of the random variable's values across possible outcomes.