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Chapter 8: Problem 5

For events \(\mathcal{A}_{1}, \ldots, \mathcal{A}_{n}\), define \(\alpha_{1}:=\mathrm{P}\left[\mathcal{A}_{1}\right],\) and for \(i=2, \ldots, n,\) define \(\alpha_{i}:=\mathrm{P}\left[\mathcal{A}_{i} \mid \mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{i-1}\right]\) (assume that \(\left.\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n-1}\right] \neq 0\right)\). Show that \(\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n}\right]=\alpha_{1} \cdots \alpha_{n}\)

Short Answer

Expert verified
Question: Show that the probability of the intersection of \(n\) events is equal to the product of their defined probabilities. Answer: To prove that the probability of the intersection of \(n\) events is equal to the product of their defined probabilities, we used the definition of conditional probability, properties of intersections of events, and induction. We showed that \(\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n}\right] = \alpha_1 \cdots \alpha_n\), where \(\alpha_i = \mathrm{P}\left[\mathcal{A}_i \mid \mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right]\) for \(i = 2, \ldots, n\) and \(\alpha_1 = \mathrm{P}\left[\mathcal{A}_1\right]\).

Step by step solution

01

Understand the given definitions

We are given a sequence of events \(\mathcal{A}_1, \ldots, \mathcal{A}_n\), and we are asked to prove that: $$\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n}\right] = \alpha_1 \cdots \alpha_n$$ where $$\begin{aligned} \alpha_1 = \mathrm{P}\left[\mathcal{A}_1\right], \\ \alpha_i = \mathrm{P}\left[\mathcal{A}_i \mid \mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right]\\ \end{aligned}$$ for \(i = 2, \ldots, n\).
02

Define conditional probability

Recall that conditional probability is defined as: $$\mathrm{P}\left[A \mid B\right] = \frac{\mathrm{P}\left[A \cap B\right]}{\mathrm{P}\left[B\right]}$$
03

Use conditional probability to connect the given definitions

We will use the definition of conditional probability to relate \(\alpha_i\) to the intersection of the events. For \(i = 2, \ldots, n\), we have: $$\alpha_i = \mathrm{P}\left[\mathcal{A}_i \mid \mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right] = \frac{\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_i\right]}{\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right]}$$
04

Calculate the probability of the intersection

To find the probability of the intersection, we will multiply both sides of the equations from Step 3 by \(\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right]\) and express the result in terms of \(\alpha_i\): $$\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_i\right] = \alpha_i \mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{i-1}\right]$$
05

Proof by induction

We will now use induction to show that \(\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n}\right] = \alpha_1 \cdots \alpha_n\). Base case: For \(n = 1\), we have: $$\mathrm{P}\left[\mathcal{A}_1\right] = \alpha_1$$ which is true by definition. Induction step: Assume that for some \(k \geq 1\), we have: $$\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{k}\right] = \alpha_1 \cdots \alpha_k$$ We want to show that: $$\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{k+1}\right] = \alpha_1 \cdots \alpha_k \alpha_{k+1}$$ Using the equation from Step 4: $$\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{k+1}\right] = \alpha_{k+1}\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{k}\right]$$ And substituting the induction assumption: $$\mathrm{P}\left[\mathcal{A}_1 \cap \cdots \cap \mathcal{A}_{k+1}\right] = \alpha_{k+1}(\alpha_1 \cdots \alpha_k) = \alpha_1 \cdots \alpha_k \alpha_{k+1}$$ Thus, the induction step holds.
06

Conclusion

By induction, we have shown that the probability of the intersection of \(n\) events is equal to the product of their defined probabilities: $$\mathrm{P}\left[\mathcal{A}_{1} \cap \cdots \cap \mathcal{A}_{n}\right] = \alpha_1 \cdots \alpha_n$$

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in probability theory that deals with the likelihood of an event occurring given that another event has already occurred. It captures the idea that the occurrence of one event can influence the probability of another. For example, if we know it's raining, the probability that people carry umbrellas is likely higher compared to a day with no rain.
In mathematical terms, the conditional probability of an event A given another event B has occurred is denoted as P(A|B), and is calculated by the formula:
\[\mathrm{P}(A \mid B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}\]
This formula tells us that to find the conditional probability, we divide the probability of the intersection of A and B by the probability of B alone. This is because when B has occurred, the sample space is reduced to only the outcomes where B is true, and we are looking at the probability of A within this reduced space. In the exercise, correctly applying conditional probability is crucial to prove the relationship between the compound probability of multiple events and their individual conditional probabilities.
Probability Theory
Probability theory is the branch of mathematics that studies random events and quantifies the likelihood of their occurrence. It provides a mathematical framework for predicting the outcomes of complex systems where chance plays a role. This theory is applied in numerous fields such as finance, insurance, psychology, and physics, to name a few.
Fundamental to this theory is the concept of an 'event,' which is a set of outcomes from a probability experiment. These events can be simple, consisting of a single outcome, or compound, involving combinations of simple events. The probability of an event is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the context of the textbook exercise, the probability theory principles are applied to derive a general product rule for the probability of the intersection of multiple events (\(\mathcal{A}_1, \ldots, \mathcal{A}_n\)).
Induction in Mathematics
Induction in mathematics is a powerful technique for proving statements, propositions, or theorems that are formulated in terms of natural numbers. It consists of two critical steps:
  • Base Case: Proving the statement is true for the initial value, often for 1 or 0.
  • Inductive Step: Assuming the statement is true for some natural number 'k' (the induction hypothesis), then proving it is also true for 'k+1'.

Through these two steps, we can effectively demonstrate the truth of the statement for all natural numbers. In our exercise, mathematical induction is used to extend the proof that the probability of the intersection of a set of events equals the product of their individual probabilities from a base case of one event to any number 'n' of events. This is done by showing that this property holds for one event (the base case), and then assuming it holds for 'k' events, it can be proved for 'k+1' events (the inductive step). This extrapolation cements the proof, establishing the relationship expressed in the problem for any number of events.

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Most popular questions from this chapter

In each step of a random walk, we toss a coin, and move either one unit to the right, or one unit to the left, depending on the outcome of the coin toss. The question is, after \(n\) steps, what is our expected distance from the starting point? Let us model this using a mutually independent family of random variables \(\left\\{Y_{i}\right\\}_{i=1}^{n},\) with each \(Y_{i}\) uniformly distributed over \(\\{-1,1\\},\) and define \(Y:=Y_{1}+\cdots+Y_{n} .\) Show that the \(c_{1} \sqrt{n} \leq \mathrm{E}[|Y|] \leq c_{2} \sqrt{n},\) for some constants \(c_{1}\) and \(c_{2}\)

Suppose \(n\) balls are thrown into \(m\) bins. Let \(\mathcal{A}\) be the event that there is some bin that is empty. Assuming that the throws are mutually independent, and that \(n \geq m(\log m+t)\) for some \(t \geq 0,\) show that \(\mathrm{P}[\mathcal{A}] \leq e^{-t}\).

Let \(f:[0,1] \rightarrow \mathbb{R}\) be a function that is "nice" in the following sense: for some constant \(c,\) we have \(|f(s)-f(t)| \leq c|s-t|\) for all \(s, t \in[0,1] .\) This condition is implied, for example, by the assumption that \(f\) has a derivative that is bounded in absolute value by \(c\) on the interval \([0,1] .\) For each positive integer \(n,\) define the polynomial \(B_{n, f}:=\sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) f(k / n) T^{k}(1-T)^{n-k} \in \mathbb{R}[T] .\) Show that \(\left|B_{n, f}(p)-f(p)\right| \leq c / 2 \sqrt{n}\) for all positive integers \(n\) and all \(p \in[0,1] .\) Hint: let \(X\) be a random variable with a binomial distribution that counts the number of successes among \(n\) Bernoulli trials, each of which succeeds with probability \(p,\) and begin by observing that \(\boldsymbol{B}_{n, f}(p)=\mathrm{E}[f(X / n)] .\) The polynomial \(\boldsymbol{B}_{n, f}\) is called the \(n\) th Bernstein approximation to \(f,\) and this result proves a classical result that any "nice" function can approximated to arbitrary precision by a polynomial of sufficiently high degree.

Let \(S\) be a set of size \(m \geq 1,\) and let \(s_{0}\) be an arbitrary, fixed element of \(S\). Let \(F\) be a random variable that is uniformly distributed over the set of all \(m^{m}\) functions from \(S\) into \(S .\) Let us define random variables \(X_{i},\) for \(i=0,1,2, \ldots,\) as follows: $$ X_{0}:=s_{0}, \quad X_{i+1}:=F\left(X_{i}\right)(i=0,1,2, \ldots) $$ Thus, the value of \(X_{i}\) is obtained by applying the function \(F\) a total of \(i\) times to the starting value \(s_{0}\). Since \(S\) has size \(m,\) the sequence \(\left\\{X_{i}\right\\}_{i=0}^{\infty}\) must repeat at some point; that is, there exists a positive integer \(n\) (with \(n \leq m\) ) such that \(X_{n}=X_{i}\) for some \(i=0, \ldots, n-1 .\) Define the random variable \(Y\) to be the smallest such value \(n .\) (a) Show that for every \(i \geq 0\) and for all \(s_{1}, \ldots, s_{i} \in S\) such that \(s_{0}, s_{1}, \ldots, s_{i}\) are distinct, the conditional distribution of \(X_{i+1}\) given the event \(\left(X_{1}=s_{1}\right) \cap\) \(\cdots \cap\left(X_{i}=s_{i}\right)\) is the uniform distribution on \(S .\) (b) Show that for every integer \(n \geq 1,\) we have \(Y \geq n\) if and only if the random variables \(X_{0}, X_{1}, \ldots, X_{n-1}\) take on distinct values. (c) From parts (a) and (b), show that for each \(n=1, \ldots, m,\) we have $$ \mathrm{P}[Y \geq n \mid Y \geq n-1]=1-(n-1) / m $$ and conclude that $$ \mathrm{P}[Y \geq n]=\prod_{i=1}^{n-1}(1-i / m) \leq e^{-n(n-1) / 2 m} $$ (d) Using part (c), show that $$ \mathrm{E}[Y]=\sum_{n \geq 1} \mathrm{P}[Y \geq n] \leq \sum_{n \geq 1} e^{-n(n-1) / 2 m}=O\left(m^{1 / 2}\right) $$ (e) Modify the above argument to show that \(\mathrm{E}[Y]=\Omega\left(m^{1 / 2}\right)\).

Show that \(\mathrm{P}[\mathcal{A} \cap B] \mathrm{P}[\mathcal{A} \cup \mathcal{B}] \leq \mathrm{P}[\mathcal{A}] \mathrm{P}[\mathcal{B}]\) for all events \(\mathcal{A}, \mathcal{B}\).
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