Problem 252 - An Introduction to Mathematical Statistics and Its Applications

Question

If \(X\) and \(Y\) have the joint pdf $$ f_{X, Y}(x, y)=2, \quad 0 \leq x

Step-by-Step Solution

Answer

The conditional probability \(P\left(0 < X < \frac{1}{2} \mid Y = \frac{3}{4}\right)\) is \( \frac{2}{3}\).

Step 1: Understand the problem and identify given values.

We are given a joint probability density function \(f_{X, Y}(x, y) = 2\) defined for \(0 \leq x < y \leq 1\). We are tasked with finding the conditional probability \(P\left(0 < X < \frac{1}{2} \mid Y = \frac{3}{4}\right)\). To find this, we first need to find the marginal density function of \(Y\).

Step 2: Find the marginal pdf of Y.

The marginal pdf of \(Y\) is obtained by integrating the joint pdf over all possible values of \(x\). Given the constraints \(0 \leq x < y \leq 1\), we get that when \(Y = \frac{3}{4}\), \(X\) ranges from 0 to \(\frac{3}{4}\). Thus, \(f_Y(y) = \int_{0}^{y} 2 dx = 2y \) for \(0 < y \leq 1\), which is 1.5 when \(Y = \frac{3}{4}\).

Step 3: Find the conditional pdf of X given Y is 3/4 and use it to find the required probability.

The conditional pdf \(f_{X|Y}(x | y)\) is the joint pdf \(f_{X, Y}(x, y)\) divided by the marginal pdf \(f_Y(y)\). When \(Y = \frac{3}{4}\), the joint pdf is 2 for \(0 \leq x < \frac{3}{4}\), so \(f_{X|Y}(\frac{1}{2}|\frac{3}{4}) = \frac{f_{X, Y}(x, y)}{f_Y(\frac{3}{4})} = \frac{2}{1.5} = \frac{4}{3}\) for \(0 < x < \frac{3}{4}\). The required probability is then \(P\left(0 < X < \frac{1}{2} \mid Y = \frac{3}{4}\right) = \int_{0}^{\frac{1}{2}} f_{X|Y}(x|\frac{3}{4}) dx = \int_{0}^{\frac{1}{2}} \frac{4}{3} dx = \frac{2}{3} \).

Key Concepts

Joint Probability Density Function

A joint probability density function (pdf) is a function used to specify the likelihood of two continuous random variables, say \(X\) and \(Y\), occurring together. It provides a complete description of the way in which these two variables interact over a specific range of their values. The joint pdf is denoted as \(f_{X,Y}(x,y)\). In our given problem, this is given by \(f_{X, Y}(x, y)=2\) for the range \(0 \leq x < y \leq 1\). This means that within this boundary, the density is constant.

Some key properties of joint pdfs to remember include:

• The non-negativity of the function: \(f_{X,Y}(x,y) \geq 0\).
• The total probability requirement: the integral over all possible values of \(x\) and \(y\) must equal 1. This confirms it is a valid pdf.
• The ability to derive marginal pdfs, which will be discussed next.

Joint probability density functions are crucial in scenarios where we need to understand the interaction or dependency between two random variables.

Marginal Probability Density Function

The marginal probability density function of a random variable is a projection of the joint pdf onto one of its axes. It is obtained by integrating the joint pdf over the range of the other variable. Essentially, it provides the pdf of a single variable, regardless of the values of the other.

For our given exercise, to find the marginal pdf of \(Y\), we integrate the joint pdf over \(X\) within the range \(0 \leq x < y\). It is expressed as:\[f_Y(y) = \int_{0}^{y} f_{X,Y}(x,y) \, dx = \int_{0}^{y} 2 \, dx = 2y\For \(0 \leq y \leq 1\)\]
This function \(f_Y(y)\) tells us the probability density of \(Y\) regardless of \(X\). Understanding marginal pdfs is crucial as they simplify our analysis to a single dimension and enable us to further analyze conditions based on one variable.

Probability Integration

Probability integration is a mathematical tool used to calculate the probability over an interval by integrating the pdf over that interval. When dealing with conditional probabilities, the integration of the conditional pdf can be essential.

For conditional scenarios, the density function is adjusted by normalizing with the marginal pdf of the conditioning event. In our problem, we calculated the conditional pdf of \(X\) given \(Y = \frac{3}{4}\) and then integrated it over the interval \((0, \frac{1}{2})\):\[P\left(0 < X < \frac{1}{2} \mid Y = \frac{3}{4}\right) = \int_{0}^{\frac{1}{2}} f_{X|Y}(x|\frac{3}{4}) dx = \int_{0}^{\frac{1}{2}} \frac{4}{3} \, dx = \frac{2}{3}\]
This integral gives us the probability of \(X\) within the specified range under the condition \(Y = \frac{3}{4}\). The concept of probability integration is fundamental, as it allows us to compute probabilities for normally distributed continuous variables and complex conditional scenarios.