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Let \(X\) and \(Y\) be independent continuous random variables with respective hazard rate functions \(\lambda_{X}(t)\) and \(\lambda_{Y}(t)\), and set \(W=\min (X, Y)\). (a) Determine the distribution function of \(W\) in terms of those of \(X\) and \(Y\). (b) Show that \(\lambda_{W}(t)\), the hazard rate function of \(W\), is given by $$ \lambda_{W}(t)=\lambda_{X}(t)+\lambda_{Y}(t) $$

Short Answer

Expert verified
The distribution function of $W$ can be found as $F_W(w) = 1 - (1 - F_X(w))(1 - F_Y(w))$. The hazard rate function of $W$, denoted by $\lambda_W(t)$, can be shown to be equal to $\lambda_X(t) + \lambda_Y(t)$.

Step by step solution

01

Define the Cumulative Distribution Function (CDF) of W

Let's define the CDF of W as F_W(w). By definition, this is the probability that W is less than or equal to w, i.e., F_W(w) = P(W ≤ w). Since W = min(X, Y), we have P(W ≤ w) = P(min(X, Y) ≤ w).
02

Use the properties of CDFs to find P(min(X, Y) ≤ w)

We will make use of the complementary probability rule: P(A) = 1 - P(A^c) where A^c is the complement of event A. Now, the complement of the event min(X, Y) ≤ w is X > w and Y > w, as both X and Y must be greater than w for min(X, Y) to be greater than w. Therefore, P(min(X, Y) ≤ w) = 1 - P(X > w, Y > w) Since X and Y are independent, we have: P(X > w, Y > w) = P(X > w)P(Y > w)
03

Compute P(X > w) and P(Y > w)

We will again make use of the complementary probability rule: P(X > w) = 1 - P(X ≤ w) = 1 - F_X(w) P(Y > w) = 1 - P(Y ≤ w) = 1 - F_Y(w) Now, substitute these back into the expression from Step 2: P(min(X, Y) ≤ w) = 1 - (1 - F_X(w))(1 - F_Y(w))
04

Obtain the CDF of W in terms of the CDFs of X and Y

From Step 3, we get: F_W(w) = 1 - (1 - F_X(w))(1 - F_Y(w)) Now, we need to determine the hazard rate function of W using the CDF of W.
05

Define the hazard rate function using the CDF

For a continuous random variable Z with CDF F_Z(z) and hazard rate function λ_Z(z): λ_Z(z) = f_Z(z) / (1 - F_Z(z)) Where f_Z(z) is the probability density function (PDF) of Z. To find the hazard rate function of W, we first need to find the PDF of W.
06

Obtain the PDF of W

Differentiate F_W(w) with respect to w to obtain the PDF, f_W(w): f_W(w) = d(F_W(w))/dw = d(1 - (1 - F_X(w))(1 - F_Y(w)))/dw f_W(w) = F_X'(w)(1 - F_Y(w)) + F_Y'(w)(1 - F_X(w))
07

Find the hazard rate function of W

Using the expression for the hazard rate function from Step 5, we now have: λ_W(w) = f_W(w) / (1 - F_W(w)) Substitute f_W(w) and F_W(w) in terms of F_X(w), F_Y(w), F_X'(w), F_Y'(w): λ_W(w) = (F_X'(w)(1 - F_Y(w)) + F_Y'(w)(1 - F_X(w))) / ((1 - F_X(w))(1 - F_Y(w))) Recall that the hazard rate functions for X and Y are λ_X(w) = F_X'(w) / (1 - F_X(w)) and λ_Y(w) = F_Y'(w) / (1 - F_Y(w)). Now we can substitute these back into λ_W(w): λ_W(w) = λ_X(w) + λ_Y(w) This completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hazard Rate Function
The hazard rate function, often referred to as the failure rate in reliability theory, represents the instantaneous rate of occurrence of an event at time \( t \), given that the event has not occurred before \( t \). It is particularly useful in fields like survival analysis and reliability engineering. For a continuous random variable \( Z \), with cumulative distribution function (CDF) \( F_Z(z) \) and probability density function (PDF) \( f_Z(z) \), the hazard rate function \( \lambda_Z(t) \) is defined as:
  • \( \lambda_Z(t) = \frac{f_Z(t)}{1 - F_Z(t)} \)
Let's break down this formula:
  • \( f_Z(t) \) is the PDF, showing the likelihood of \( Z \) taking a value close to \( t \).
  • \( 1 - F_Z(t) \) represents the probability that \( Z \) is greater than \( t \), indicating the "survival" past time \( t \).
In simple words, the hazard rate is the chance per unit time that the event will happen right after \( t \), given it didn't happen before. This function helps differentiate between different types of distributions based on how likely an event is to occur at a given moment.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a random variable gives us the probability that the variable will take a value less than or equal to a certain number. For a random variable \( X \), the CDF is noted as \( F_X(x) \), and it reflects the cumulative probability up to the point \( x \). The formula is:
  • \( F_X(x) = P(X \leq x) \)
The CDF is crucial because it encompasses all the probabilities of outcomes, effectively summarizing the distribution of a random variable. Moreover, the CDF is a non-decreasing function, starting at 0 and eventually reaching 1 as \( x \) approaches infinity.

When working with two independent random variables, such as in our exercise where \( W = \min(X, Y) \), the CDF of \( W \) is derived using the relation between \( X \) and \( Y \). Specifically, for such a scenario, the CDF of \( W \), \( F_W(w) \), is computed as:
  • \( F_W(w) = 1 - (1 - F_X(w))(1 - F_Y(w)) \)
This formula arises by considering the complementary probabilities of both \( X \) and \( Y \) being greater than \( w \). It's a clear example of how the concept of minimum translates into probability terms for cumulative events.
Independent Random Variables
Two random variables are independent when the occurrence of one does not affect the probability of occurrence of another. In probabilistic terms, this means that knowing the outcome of one random variable gives no information about the other. Mathematically, two random variables \( X \) and \( Y \) are independent if:
  • \( P(X \leq x, Y \leq y) = P(X \leq x) \cdot P(Y \leq y) \)
This property of independence simplifies many calculations, such as determining joint probabilities, as shown in the exercise:
  • \( P(X > w, Y > w) = P(X > w) \cdot P(Y > w) \)
In the context of the exercise, this independence is crucial in the calculation of the CDF of \( W = \min(X, Y) \) and in determining the hazard rate function. It allows us to separately consider each variable's contribution to the overall behavior of \( W \). The independence leads directly to straightforward computation for functions like minimum or maximum, where each event's probability factors without interference from the others.

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Most popular questions from this chapter

Suppose that \(F(x)\) is a cumulative distribution function. Show that (a) \(F^{\prime \prime}(x)\) and (b) \(1-[1-F(x)]^{n}\) are also cumulative distribution functions when \(n\) is a positive integer. HINT: Let \(X_{1}, \ldots, X_{n}\) be independent random variables having the common distribution function \(\vec{F}\). Define random variables \(Y\) and \(Z\) in terms of the \(X\) so that \(P\\{Y \leq x\\}=F^{n}(x)\), and \(P\\{Z \leq x\\}=1-[1-F(x)]^{n}\).

Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of \(n\) independent uniform \((0,1)\) random variables. Prove that for \(1 \leq k \leq n+1\), $$ P\left\\{X_{(k)}-X_{(k-1)}>t\right\\}=(1-t)^{n} $$ where \(X_{0} \equiv 0, X_{n+1} \equiv t\).

Each throw of a unfair die lands on each of the odd numbers \(1,3,5\) with probability \(C_{3}\) and on each of the even numbers with probability \(2 C\). (a) Find \(C\). (b) Suppose that the die is tossed. Let \(X\) equal 1 if the result is an even -number, and let it be 0 otherwise. Also, let \(Y\) equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of \(X\) and \(Y\). Suppose now that 12 independent tosses of the die are made. e(c) Find the probability that each of the six outcomes occurs exactly twice. (d) Find the probability that 4 of the outcomes are either one or two, 4 are either three or four, and 4 are either five or six. (e) Find the probability that at least 8 of the tosses land on even numbers.

Solve Buffon's needle problem when \(L>D\). ANSWER: \(\frac{2 L}{\pi D}(1-\sin \theta)+2 \theta / \pi\), where \(\theta\) is such that \(\cos \theta=D / L\).

The following dartboard is a square whose sides are of length 6 . The three circles are all centered at the center of the board and are of radii 1,2 , and 3. Darts landing within the circle of radius 1 score 30 points, those landing outside this circle but within the circle of radius 2 are worth 20 points, and those landing outside the circle of radius 2 but within the circle of radius 3 are worth 10 points. Darts that do not land within the circle of radius 3 do not score any points. Assuming that each dart that you throw will, independent of what occurred on your previous throws, land on a point uniformly distributed in the square, find the probabilities of the following events. (a) You score 20 on a throw of the dart. (b). You score at least 20 on a throw of the dart. (c) You score 0 on a throw of the dart. (d) The expected value of your score on a throw of the dart. (e) Both of your first two throws score at least 10 . (f) Your total score after two throws is 30 .

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