Finding Constant C in Joint Probability Density Function

To find the value of the constant C in a given joint probability density function (PDF), \( f_{XY}(x, y) \), we must use the fundamental property that the total probability over the entire sample space is equal to 1. For a continuous joint PDF defined over a region R, this property is expressed by the integral:

\[ \iint_R f_{XY}(x, y) \, dA = 1 \]

In this question, the joint PDF is given by \( f_{XY}(x, y) = x^2 + Cy \) and it is defined over the rectangular region \( 0 \le x \le 1 \) and \( 0 \le y \le 1 \). Therefore, the integral becomes:

\[ \int_{0}^{1} \int_{0}^{1} (x^2 + Cy) \, dy \, dx = 1 \]

We need to evaluate this double integral step-by-step.

Step 1: Integrate with respect to y

We first integrate the function \( x^2 + Cy \) with respect to y, treating x as a constant, from y = 0 to y = 1:

\[ \int_{0}^{1} (x^2 + Cy) \, dy = \left[ x^2y + C\frac{y^2}{2} \right]_{y=0}^{y=1} \]

Now, we evaluate the expression at the limits of integration:

\[ = \left( x^2(1) + C\frac{1^2}{2} \right) - \left( x^2(0) + C\frac{0^2}{2} \right) \]

\[ = \left( x^2 + \frac{C}{2} \right) - (0 + 0) \]

\[ = x^2 + \frac{C}{2} \]

Step 2: Integrate the result with respect to x

Next, we integrate the result from Step 1, which is \( x^2 + \frac{C}{2} \), with respect to x, from x = 0 to x = 1:

\[ \int_{0}^{1} \left( x^2 + \frac{C}{2} \right) \, dx = \left[ \frac{x^3}{3} + \frac{C}{2}x \right]_{x=0}^{x=1} \]

Now, we evaluate this expression at the limits of integration:

\[ = \left( \frac{1^3}{3} + \frac{C}{2}(1) \right) - \left( \frac{0^3}{3} + \frac{C}{2}(0) \right) \]

\[ = \left( \frac{1}{3} + \frac{C}{2} \right) - (0 + 0) \]

\[ = \frac{1}{3} + \frac{C}{2} \]

Step 3: Set the integral result equal to 1 and solve for C

According to the property of joint probability density functions, the total integral must equal 1:

\[ \frac{1}{3} + \frac{C}{2} = 1 \]

Now, we solve this linear equation for C:

Subtract \( \frac{1}{3} \) from both sides:

\[ \frac{C}{2} = 1 - \frac{1}{3} \]

\[ \frac{C}{2} = \frac{3}{3} - \frac{1}{3} \]

\[ \frac{C}{2} = \frac{2}{3} \]

Multiply both sides by 2:

\[ C = 2 \times \frac{2}{3} \]

\[ C = \frac{4}{3} \]

Thus, the value of the constant C that makes \( f_{XY}(x, y) = x^2 + Cy \) a valid joint probability density function over the specified region is \( \frac{4}{3} \).

Revision Table: Joint PDF Constant Calculation

Additional Information: Joint Probability Density Functions

A joint probability density function, \( f_{XY}(x, y) \), provides the probability distribution for two continuous random variables, X and Y. For \( f_{XY}(x, y) \) to be a valid joint PDF, it must satisfy two conditions:

• Non-negativity: \( f_{XY}(x, y) \ge 0 \) for all possible values of x and y.
• Normalization: The total volume under the surface \( z = f_{XY}(x, y) \) over the entire range of X and Y must be equal to 1. This is represented by the double integral \( \iint_R f_{XY}(x, y) \, dA = 1 \), where R is the region where \( f_{XY}(x, y) \) is defined.

Calculating the probability that the random variables (X, Y) fall within a specific subregion D of the sample space R is done by integrating the joint PDF over that subregion D:

\[ P((X, Y) \in D) = \iint_D f_{XY}(x, y) \, dA \]

Marginal PDFs for X and Y can be obtained by integrating the joint PDF over the entire range of the other variable:

• Marginal PDF of X: \( f_X(x) = \int_{-\infty}^{\infty} f_{XY}(x, y) \, dy \)
• Marginal PDF of Y: \( f_Y(y) = \int_{-\infty}^{\infty} f_{XY}(x, y) \, dx \)

Understanding these properties is crucial when working with joint probability distributions and problems involving finding unknown constants or calculating probabilities.