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In probability, the normalization condition ensures that a joint probability density function (pdf) represents a valid probability distribution. For any joint pdf, the total probability over its entire range must be equal to 1. This means if you sum up all the possible outcomes in a probability space, they represent certainty—a 100% chance of something within that space occurring. To satisfy this condition for the function \(f(x, y) = cxy\) that is defined for \(0 < x < 3\) and \(0 < y < 3\), we have to integrate the function over all possible values of \(x\) and \(y\) to ensure it equals 1.The process involves two integrations:

• First, integrate with respect to \(y\): \(\int_{0}^{3} xy \, dy\) gives \(\frac{9x}{2}\).
• Next, integrate \(\frac{9x}{2}\) with respect to \(x\): \(\int_{0}^{3} \frac{9x}{2} \, dx\) results in \(\frac{81}{4}\).

By equating \(c \cdot \frac{81}{4} = 1\), we find \(c = \frac{4}{81}\). This normalization ensures that \(f(x, y) = cxy\) is indeed a valid joint pdf over its specified range.

The marginal probability distribution helps you understand the distribution of a subset of variables within a joint probability framework. In simpler terms, it shows the probability of one variable, irrespective of the other, in a multivariate distribution. For example, if you have a joint distribution \(f(x, y)\), the marginal probability distribution of \(X\) is found by integrating out \(Y\). To find this, you integrate the joint pdf \(f(x, y)\) over all possible values of \(y\). Here’s the process:

• Start with the joint pdf \(f(x, y) = \frac{4}{81}xy\).
• Integrate over \(y\) from 0 to 3: \(\int_{0}^{3} \frac{4}{81}xy \, dy\).

This simplifies to \(\frac{4x}{81} \times \frac{9}{2} = \frac{2x}{27}\). Thus, the marginal probability distribution of \(X\) is \(\frac{2x}{27}\), indicating how the random variable \(X\) behaves irrespective of \(Y\).

The expected value provides the average or mean outcome of a random variable over a large number of trials. It’s a fundamental concept typically denoted by \(E(X)\) for random variable \(X\). For a joint probability density function, finding the expected value of one variable requires integrating the product of that variable and the joint pdf over the range of possible values of both variables.In our example with the joint pdf \(f(x, y) = \frac{4}{81}xy\), we determine \(E(X)\) by using the formula:\[ E(X) = \int_{0}^{3} \int_{0}^{3} x \cdot \frac{4}{81}xy \, dy \, dx \]This involves:

• Integrating \(x^2y\) with respect to \(y\) first, which gives \(\frac{x^2 \cdot 9}{2}\).
• Then, integrating the result with respect to \(x\), gives the final expected value \(E(X)\).

Through these steps, the expected value ultimately represents the center of mass for the random variable \(X\) under the probability distribution defined by the joint pdf.