If X1 And X2 Are Independent Nonnegative Continuous Random Variables, Show That P{X1 < X2 | Min(X1, X2)}
Understanding the behavior of random variables and their relationships is fundamental in probability theory and statistics. Specifically, analyzing the conditional probability involving two independent nonnegative continuous random variables, X1 and X2, provides insight into their comparative likelihoods. The problem asks us to demonstrate that, given the minimum of the two variables, the probability that X1 is less than X2 can be expressed in a particular form. This article thoroughly explores this problem, providing detailed explanations, step-by-step derivations, and relevant concepts necessary for a comprehensive understanding.
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Introduction to the Problem
The statement involves two key random variables:
- X1 and X2: Independent, nonnegative, continuous random variables.
- Goal: To derive and understand the expression for the conditional probability \( P\{X1 < X2 \mid \min(X1, X2) \} \).
In probability, the notation indicates the probability that X1 is less than X2, given the value of their minimum. Since X1 and X2 are independent and nonnegative, their joint distribution has specific properties that facilitate this derivation.
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Preliminaries and Definitions
Before delving into the derivation, it's essential to clarify some foundational concepts:
1. Independence of Random Variables
Two random variables X and Y are independent if their joint probability density function (pdf) factors into the product of their marginal pdfs:
\[
f{X,Y}(x,y) = fX(x) \times f_Y(y),
\]
for all x, y in their respective support.
2. Nonnegative Continuous Random Variables
X1 and X2 are nonnegative, meaning:
\[
P\{X_i \geq 0\} = 1, \quad i=1,2.
\]
Their pdfs are defined over [0, ∞):
\[
f{Xi}(x), \quad x \geq 0.
\]
3. Minimum of Two Random Variables
\[
\min(X1, X2) = Z,
\]
which is itself a random variable. Its distribution depends on the distributions of X1 and X2.
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Deriving the Conditional Probability
The main aim is to compute:
\[
P\{X1 < X2 \mid \min(X1, X2) = z\}.
\]
This involves understanding the joint distribution of X1 and X2, conditioned on their minimum being a specific value z.
Step 1: Express the Conditional Probability
By the definition of conditional probability:
\[
P\{X1 < X2 \mid \min(X1, X2) = z\} = \frac{P\{X1 < X2, \min(X1, X2) = z\}}{f_{\min}(z)},
\]
where \(f{\min}(z)\) is the pdf of \(\min(X1, X_2)\).
Since the minimum equals z, the joint behavior of X1 and X2 must satisfy:
\[
\min(X1, X2) = z,
\]
and both are ≥ z.
The numerator involves the joint probability that:
- \(\min(X1, X2) = z\),
- and \(X1 < X2\).
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Step 2: Understanding the Event \(\{X1 < X2, \min(X1, X2) = z\}\)
For the minimum to be z, at least one of the variables must be equal to z, and the other must be greater than or equal to z.
The event \(X1 < X2\) and \(\min(X1, X2) = z\) can be partitioned into two disjoint cases:
- Case 1: \(X1 = z\) and \(X2 > z\),
- Case 2: \(X1 > z\) and \(X2 = z\).
Since in the first case, \(X1 = z\) and \(X2 > z\), and in the second, \(X2 = z\) and \(X1 > z\), for the event \(X1 < X2\) to happen, only the first case contributes because:
- If \(X2 = z\) and \(X1 > z\), then \(X1 > X2\), which is not consistent with \(X1 < X2\).
Therefore, the relevant event is:
\[
\{X1 = z, X2 > z\}.
\]
And, similarly, for the denominator, the distribution of \(\min(X1, X2)\), we need the joint pdf over the region where:
\[
X1 = z, \quad X2 \geq z,
\]
and vice versa, but since only the case where \(X1 = z\) and \(X2 > z\) contributes to the numerator, we focus on that.
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Step 3: Expressing the Numerator
The probability that:
- \(X_1 = z\),
- \(X_2 > z\),
- and \(X1 < X2\).
Given the independence and continuous nature, the probability that \(X_1 = z\) is zero (since continuous variables have zero probability at a specific point). However, the joint probability density on the boundary is described via the joint pdf evaluated at the boundary.
To handle this, we consider the joint probability density functions over infinitesimal intervals:
\[
P\{X1 \in [z, z + dz], X2 > z\} \approx f{X1}(z) dz \times P\{X2 > z\} = f{X1}(z) dz \times (1 - F{X_2}(z)),
\]
where \(F{X2}(z)\) is the CDF of \(X_2\).
Similarly, for the joint distribution, the probability that \(X1 < X2\), with \(X1\) near z and \(X2\) greater than z, is:
\[
\int{x2=z}^{\infty} f{X1}(z) \times f{X2}(x2) dx2,
\]
since independence implies the joint pdf factorizes.
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Step 4: Formal Expression for the Conditional Probability
Putting it all together, the conditional probability becomes:
\[
P\{X1 < X2 \mid \min(X1, X2) = z\} = \frac{P\{X1 = z, X2 > z, X1 < X2\}}{f_{\min}(z)}.
\]
But as the probability of the variables taking exact values is zero, we use the density functions:
\[
P\{X1 < X2 \mid \min(X1, X2) = z\} = \frac{f{X1}(z) \int{x2=z}^{\infty} f{X2}(x2) dx2}{f_{\min}(z)}.
\]
Similarly, the pdf of the minimum, \(f_{\min}(z)\), is known for independent variables:
\[
f{\min}(z) = f{X1}(z) [1 - F{X2}(z)] + f{X2}(z) [1 - F{X_1}(z)].
\]
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Final Result and Interpretation
Using the above, the conditional probability simplifies to:
\[
\boxed{
P\{X1 < X2 \mid \min(X1, X2) = z\} = \frac{f{X1}(z) [1 - F{X2}(z)]}{f{X1}(z) [1 - F{X2}(z)] + f{X2}(z) [1 - F{X1}(z)]}.
}
\]
This expression indicates that, given the minimum of the two variables is z, the probability that X1 is less than X2 depends on the relative densities and tail probabilities of the two distributions at z.
---
Special Cases and Intuition
- Identical Distributions: If \(X1\) and \(X2\) are identically distributed, then \(f{X1}(z) = f{X2}(z)\) and \(F{X1}(z) = F{X2}(z)\). The expression reduces to:
