Problem 81 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) is a continuous random variable with probability distribution $$ f_{X}(x)=\frac{x}{18}, \quad 0 \leq x \leq 6 $$ (a) Determine the probability distribution of the random variable \(Y=2 X+10\). (b) Determine the expected value of \(Y\).

Step-by-Step Solution

Answer

The probability distribution of \(Y\) is \(f_Y(y) = \frac{y-10}{72}\) for \(10 \leq y \leq 22\), and the expected value of \(Y\) is 18.

Step 1: Identify the Transformation

We have the transformation function for the random variable from \(X\) to \(Y = 2X + 10\).

Step 2: Find the Range of Y

Since \(X\) ranges from 0 to 6, we need to find how this transforms to \(Y\). The minimum value of \(Y\) occurs when \(X = 0\), which gives \(Y = 2(0) + 10 = 10\). The maximum value of \(Y\) occurs when \(X = 6\), giving \(Y = 2(6) + 10 = 22\). So, the range of \(Y\) is \(10 \leq Y \leq 22\).

Step 3: Determine the Probability Density Function of Y

To find the distribution of \(Y\), we use the transformation of variables technique. The relation between \(X\) and \(Y\) is \(Y = 2X + 10\), so \(X = \frac{Y - 10}{2}\). The probability density function of \(Y\) can be determined by substituting \(X\) in terms of \(Y\) into the density function of \(X\): \[ f_Y(y) = f_X\left(\frac{y-10}{2}\right) \left| \frac{d}{dy}\left(\frac{y-10}{2}\right) \right| = \frac{\frac{y-10}{2}}{18} \times \frac{1}{2} = \frac{y-10}{72}, \quad 10 \leq y \leq 22.\]

Step 4: Calculate the Expected Value of X

The expected value of a random variable \(X\) can be calculated by the integral of \(x\) times its probability density function over its range: \[E[X] = \int_{0}^{6} x \cdot \frac{x}{18} \, dx = \frac{1}{18} \int_{0}^{6} x^2 \, dx.\]Evaluating the integral, we have: \[\frac{1}{18} \left[ \frac{x^3}{3} \right]_{0}^{6} = \frac{1}{18} \left( \frac{6^3}{3} - \frac{0^3}{3} \right) = \frac{1}{18} \cdot 72 = 4.\]

Step 5: Calculate the Expected Value of Y

Using the linear transformation property of the expected value \(E[aX + b] = aE[X] + b\), we find: \[E[Y] = E[2X + 10] = 2E[X] + 10 = 2 \times 4 + 10 = 18.\]

Key Concepts

Random Variable Transformation

Transforming a random variable entails modifying it using a specific mathematical function. This is often done to obtain a new variable representing some adjusted version of the original. Consider the exercise where we have a transformation applied to the continuous random variable \(X\). The function given is \(Y = 2X + 10\). This is a linear transformation because the relation involves scaling \(X\) by a constant factor and then adding a constant. Such transformations are common because they maintain the structure of probability distributions while allowing for changes in scale or location of the variables.

When transforming a variable, it's crucial to determine the new limits for the transformed variable. If \(X\) ranges between 0 and 6, then \(Y\) will have a range derived by plugging the extremities of \(X\) into the transformation formula. This leads to \(Y\) spanning from 10 to 22, as calculated by substituting the minimum and maximum values of \(X\) into the equation \(Y = 2X + 10\). It is crucial to understand this procedure to properly model real-world problems where changes in measurement or scale might be required.

Probability Density Function

The probability density function (PDF) is a vital concept for continuous random variables, representing how the probability of different outcomes is distributed over values. For a random variable \(X\) with a known PDF \(f_X(x)\), transformation affects how probabilities are distributed over the new variable \(Y\).

In our example, if the original PDF of \(X\) is \(f_X(x) = \frac{x}{18}\), the PDF for \(Y\) is determined by changing variables via the given transformation \(Y = 2X + 10\), or equivalently \(X = \frac{Y - 10}{2}\). The transformed PDF, \(f_Y(y)\), is derived by substituting \(X\) in terms of \(Y\) into \(f_X(x)\), and adjusting for the change made in scaling, typically involving a derivative for continuous cases, as shown:

• Substitute \(X\) into \(f_X(x)\): \(f_X\left(\frac{y-10}{2}\right)\).
• Adjust for the transformation's scale by applying the derivative, yielding: \(f_Y(y) = \frac{y-10}{72}\).

Understanding these steps is essential for anyone working with transformed random variables, as it dictates how the original distribution is reshaped.

Expected Value Calculation

The expected value, or mean, is a central measure in probability helping to gauge the average outcome. To compute the expected value of a transformed variable like \(Y\), one can rely on both direct integration and linear properties of expectation.

When calculating directly for \(X\), the expected value \(E[X]\) is found via integration:\[E[X] = \int_{0}^{6} x \cdot \frac{x}{18} \, dx = 4.\]For a linear transformation \(Y = 2X + 10\), the expected value \(E[Y]\) can be efficiently computed using linearity:\[E[Y] = E[2X + 10] = 2E[X] + 10.\]This property simplifies calculations significantly because you're leveraging the expected value of the original variable \(X\) without needing a new integration for \(Y\). Hence, the expected value \(E[Y] = 18\).

These intuitive steps demonstrate that understanding transformation properties and mathematical operations like integration can simplify and illuminate the process of expected value computation for continuous random variables.