Problem 253
Question
Find \(P\left(X<1 \mid Y=1 \frac{1}{2}\right)\) if \(X\) and \(Y\) have the joint
pdf
$$
f_{X, Y}(x, y)=x y / 2, \quad 0 \leq x
Step-by-Step Solution
Verified Answer
The conditional probability \(P(X<1 | Y=1\frac{1}{2}) = \frac{4}{7}\)
1Step 1: Set up the integral for the conditional probability
2Step 2: Find conditional pdf and substitute into integral
To find \(f_{X \mid Y=1\frac{1}{2}}(x)\), substitute \(Y = 1\frac{1}{2}\) into \(f_{X,Y}(x,y)\) and renormalize by dividing by the marginal pdf of Y: \[f_{X \mid Y=1\frac{1}{2}}(x) = \frac{f_{X,Y}(x,1\frac{1}{2})}{\int_{0}^{1\frac{1}{2}} f_{X,Y}(x,1\frac{1}{2}) \,dx} = \frac{x(1\frac{1}{2})/2}{\int_{0}^{1\frac{1}{2}} (x(1\frac{1}{2})/2) \,dx} = \frac{2x}{1\frac{3}{4}-0} = \frac{8}{7}x\] Thus we have: \[P\left(X<1 \mid Y=1 \frac{1}{2}\right) = \int_{0}^{1} \frac{8}{7}x \,dx \]
3Step 3: Evaluate the integral
The integral can be evaluated using the power rule for integrals: \[ \int x^n \, dx = \frac{1}{n+1} x^{n+1} \] Thus, we get: \[P\left(X<1 \mid Y=1 \frac{1}{2}\right) = \left[\frac{8}{14} x^2 \right]_0^1 = \frac{8}{14} - 0 = \frac{4}{7}\]
Key Concepts
Understanding Joint Probability Density FunctionExploring Marginal DistributionConditional Probability Distribution DemystifiedThe Role of Integration in Probability
Understanding Joint Probability Density Function
The joint probability density function, often abbreviated as joint pdf, is a crucial concept in understanding the relationships between two continuous random variables. In probability theory, it defines how likely each pair of outcomes is to occur together. For two random variables, say \(X\) and \(Y\), the joint pdf is denoted as \(f_{X,Y}(x,y)\). This function provides a comprehensive layout of the probability distribution over the range of possible \((x, y)\) pairs.
- For example, consider the joint pdf \(f_{X,Y}(x,y) = \frac{xy}{2}\) defined over the region \(0 \leq x < y \leq 2\). This tells us how the probability is distributed for the pairs \((x, y)\) within the specified limits.
- It's important to note that for a joint pdf, the integral of the joint pdf over all possible values of \(x\) and \(y\) must equal 1. This ensures that the total probability is distributed throughout the space correctly.
By understanding the joint pdf, one can analyze how two variables interact and predict various probabilistic outcomes.
- For example, consider the joint pdf \(f_{X,Y}(x,y) = \frac{xy}{2}\) defined over the region \(0 \leq x < y \leq 2\). This tells us how the probability is distributed for the pairs \((x, y)\) within the specified limits.
- It's important to note that for a joint pdf, the integral of the joint pdf over all possible values of \(x\) and \(y\) must equal 1. This ensures that the total probability is distributed throughout the space correctly.
By understanding the joint pdf, one can analyze how two variables interact and predict various probabilistic outcomes.
Exploring Marginal Distribution
Marginal distribution helps us understand the probability of a single random variable within the context of a joint distribution. When dealing with multiple variables, marginalization allows us to focus on one variable, effectively summarizing the joint distribution.
- Marginal pdf can be found by integrating the joint pdf with respect to the other variable. This effectively "sums up" the probabilities of the undesired variable.
- For instance, if we have the joint pdf \(f_{X,Y}(x,y) = \frac{xy}{2}\), the marginal probability density of \(Y\) is obtained by integrating over \(X\): \[f_Y(y) = \int_{0}^{y} f_{X,Y}(x,y) \, dx\]
- Doing this, we find the probability distribution of \(Y\) itself, independent of \(X\).
Marginal distribution is fundamental in probabilistic analysis as it aids in understanding the behavior of individual variables in a multivariable system.
- Marginal pdf can be found by integrating the joint pdf with respect to the other variable. This effectively "sums up" the probabilities of the undesired variable.
- For instance, if we have the joint pdf \(f_{X,Y}(x,y) = \frac{xy}{2}\), the marginal probability density of \(Y\) is obtained by integrating over \(X\): \[f_Y(y) = \int_{0}^{y} f_{X,Y}(x,y) \, dx\]
- Doing this, we find the probability distribution of \(Y\) itself, independent of \(X\).
Marginal distribution is fundamental in probabilistic analysis as it aids in understanding the behavior of individual variables in a multivariable system.
Conditional Probability Distribution Demystified
Conditional probability distribution gives us the probability of one variable occurring given that another variable is known. Essentially, it slices the joint distribution, providing insights into one variable based on the occurrence of another.
- To determine the conditional probability distribution \(f_{X \mid Y}(x \mid y)\), we divide the joint pdf by the marginal pdf of the given variable \(Y\).
- In our exercise, to find \(P(X<1 \mid Y=1\frac{1}{2})\), the conditional pdf \(f_{X \mid Y=1\frac{1}{2}}(x)\) is obtained as:
\[f_{X \mid Y=1\frac{1}{2}}(x) = \frac{f_{X,Y}(x,1\frac{1}{2})}{f_Y(1\frac{1}{2})}\]
- This calculation gives finer detail on how \(X\) behaves when \(Y\) is known specific to \(Y = 1\frac{1}{2}\).
Understanding conditional distributions is essential as it offers nuanced insights into dependencies between variables.
- To determine the conditional probability distribution \(f_{X \mid Y}(x \mid y)\), we divide the joint pdf by the marginal pdf of the given variable \(Y\).
- In our exercise, to find \(P(X<1 \mid Y=1\frac{1}{2})\), the conditional pdf \(f_{X \mid Y=1\frac{1}{2}}(x)\) is obtained as:
\[f_{X \mid Y=1\frac{1}{2}}(x) = \frac{f_{X,Y}(x,1\frac{1}{2})}{f_Y(1\frac{1}{2})}\]
- This calculation gives finer detail on how \(X\) behaves when \(Y\) is known specific to \(Y = 1\frac{1}{2}\).
Understanding conditional distributions is essential as it offers nuanced insights into dependencies between variables.
The Role of Integration in Probability
Integration in probability serves as the bridge between finding distributions and calculating probabilities. It transforms probability findings into numerical insights.
- Integrating the joint pdf over a specific region gives the probability that the variables fall within that region.
- For the problem, to find the probability \(P(X < 1 \mid Y = 1\frac{1}{2})\), integration is used to evaluate: \[\int_{0}^{1} f_{X \mid Y=1\frac{1}{2}}(x) \, dx\]
- This integration from 0 to 1 over the conditional pdf calculates the total probability for \(X\) given \(Y\).- The process of integration allows us to derive cumulative probabilities, moments, and expectations, which are vital in statistical analysis.
By mastering integration techniques, one significantly enhances their ability to interpret and solve a wide range of probability problems.
- Integrating the joint pdf over a specific region gives the probability that the variables fall within that region.
- For the problem, to find the probability \(P(X < 1 \mid Y = 1\frac{1}{2})\), integration is used to evaluate: \[\int_{0}^{1} f_{X \mid Y=1\frac{1}{2}}(x) \, dx\]
- This integration from 0 to 1 over the conditional pdf calculates the total probability for \(X\) given \(Y\).- The process of integration allows us to derive cumulative probabilities, moments, and expectations, which are vital in statistical analysis.
By mastering integration techniques, one significantly enhances their ability to interpret and solve a wide range of probability problems.
Other exercises in this chapter
Problem 251
Suppose that \(X\) and \(Y\) are distributed according to the joint pdf $$ f_{X, Y}(x, y)=\frac{2}{5} \cdot(2 x+3 y), \quad 0 \leq x \leq 1, \quad 0 \leq y \leq
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If \(X\) and \(Y\) have the joint pdf $$ f_{X, Y}(x, y)=2, \quad 0 \leq x
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Suppose that \(X_{1}, X_{2}, X_{3}, X_{4}\), and \(X_{5}\) have the joint pdf $$ f x_{1}, X_{2}, X_{3}, X_{4}, X_{5}\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\righ
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Suppose the random variables \(X\) and \(Y\) are jointly distributed according to the pdf $$ f_{X, Y}(x, y)=\frac{6}{7}\left(x^{2}+\frac{x y}{2}\right), \quad 0
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