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Chapter 8: Problem 13

Let \(\left\\{X_{i}\right\\}_{i=1}^{n}\) be a family of random variables, and let \(S_{i}\) be the image of \(X_{i}\) for \(i=1, \ldots, n .\) Show that \(\left\\{X_{i}\right\\}_{i=1}^{n}\) is mutually independent if and only if for each \(i=2, \ldots, n,\) and for all \(s_{1} \in S_{1}, \ldots, s_{i} \in S_{i},\) we have $$ \mathrm{P}\left[X_{i}=s_{i} \mid\left(X_{1}=s_{1}\right) \cap \cdots \cap\left(X_{i-1}=s_{i-1}\right)\right]=\mathrm{P}\left[X_{i}=s_{i}\right] $$

Short Answer

Expert verified
Question: Prove that a family of random variables \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent if and only if the probability of each \(X_{i}\) conditioned on the values of the previous random variables is equal to the probability of \(X_{i}\) for each \(i=2,\ldots,n\). Answer: We prove both directions: 1. If \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent, the probability of each \(X_{i}\) conditioned on the values of the previous random variables is equal to the probability of \(X_{i}\) for each \(i=2,\ldots,n\). 2. If the probability of each \(X_{i}\) conditioned on the values of the previous random variables is equal to the probability of \(X_{i}\) for each \(i=2,\ldots,n\), then \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent. Using the definition of mutual independence, we find that the given condition holds under these assumptions. Using induction on the number of random variables, we can show that the random variables are mutually independent if the given condition holds.

Step by step solution

01

Definition of Mutual Independence

A family of random variables \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent if for any subset \(I\subseteq \left\{1,\ldots, n\right\}\) and any choice of \(s_i \in S_i\) for \(i\in I\), we have $$ \mathrm{P}\left[\bigcap_{i\in I}\left(X_{i}=s_{i}\right)\right]=\prod_{i\in I}\mathrm{P}\left[X_{i}=s_{i}\right]. $$
02

Prove the forward implication (Mutual independence implies the given condition)

Assume \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent. For each \(i=2,\ldots,n\), we want to show that the given condition holds. Consider the conditional probability for \(i\): $$ \mathrm{P}\left[X_{i}=s_{i} \mid\left(X_{1}=s_{1}\right) \cap \cdots \cap\left(X_{i-1}=s_{i-1}\right)\right]. $$ By the definition of conditional probability, we have: $$ \mathrm{P}\left[X_{i}=s_{i} \mid\left(X_{1}=s_{1}\right) \cap \cdots \cap\left(X_{i-1}=s_{i-1}\right)\right]=\frac{\mathrm{P}\left[\left(X_{i}=s_{i}\right) \cap \left(X_{1}=s_{1}\right) \cap \cdots \cap\left(X_{i-1}=s_{i-1}\right)\right]}{\mathrm{P}\left[\left(X_{1}=s_{1}\right) \cap \cdots \cap\left(X_{i-1}=s_{i-1}\right)\right]}. $$ Since the random variables are mutually independent, we can apply the mutual independence definition in the numerator and denominator, yielding: $$ =\frac{\mathrm{P}\left[X_{i}=s_{i}\right]\prod_{j=1}^{i-1}\mathrm{P}\left[X_{j}=s_{j}\right]}{\prod_{j=1}^{i-1}\mathrm{P}\left[X_{j}=s_{j}\right]}=\mathrm{P}\left[X_{i}=s_{i}\right]. $$ This proves that if \(\left\{X_{i}\right\}_{i=1}^{n}\) is mutually independent, then the given condition holds.
03

Prove the reverse implication (Given condition implies mutual independence)

Assume the given condition holds for each \(i=2,\ldots,n\). We want to show that the random variables are mutually independent. Consider an arbitrary subset \(I\subseteq \left\{1,\ldots, n\right\}\) and an arbitrary choice of \(s_i \in S_i\) for \(i\in I\). Let \(I=\left\{i_1,\cdots,i_k\right\}\), where \(1\le i_1< \cdots < i_k \le n\). We will use induction on \(k\) to prove that $$ \mathrm{P}\left[\bigcap_{j=1}^k\left(X_{i_j}=s_{i_j}\right)\right]=\prod_{j=1}^k\mathrm{P}\left[X_{i_j}=s_{i_j}\right]. $$ - Base case \((k=1)\): When \(k=1\), this is true by the definition of probability. - Inductive step: Assume the statement holds for \(k-1\). For \(k\), we have: $$ \mathrm{P}\left[\bigcap_{j=1}^k\left(X_{i_j}=s_{i_j}\right)\right]=\mathrm{P}\left[\left(X_{i_k}=s_{i_k}\right) \cap \left(\bigcap_{j=1}^{k-1}\left(X_{i_j}=s_{i_j}\right)\right)\right]. $$ Now, we can apply the given condition for \(i=i_k\). Notice that \(X_{i_k}=s_{i_k}\) only depends on the previous random variables, which are all in the intersection in the conditioning term. Therefore, we have: $$ =\mathrm{P}\left[X_{i_k}=s_{i_k}\right]\mathrm{P}\left[\bigcap_{j=1}^{k-1}\left(X_{i_j}=s_{i_j}\right) \mid \left(X_{i_k}=s_{i_k}\right)\right]. $$ By the induction hypothesis for \(k-1\): $$ =\mathrm{P}\left[X_{i_k}=s_{i_k}\right]\mathrm{P}\left[\bigcap_{j=1}^{k-1}\left(X_{i_j}=s_{i_j}\right)\right]=\prod_{j=1}^k\mathrm{P}\left[X_{i_j}=s_{i_j}\right]. $$ Thus, the induction is complete, and the random variables are mutually independent. This concludes the proof for both directions of the condition.

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Most popular questions from this chapter

This exercise develops an alternative proof, based on probability theory, of Theorem 2.11 . Let \(n\) be a positive integer and consider an experiment in which a number \(a\) is chosen uniformly at random from \(\\{0, \ldots, n-1\\}\). If \(n=p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\) is the prime factorization of \(n,\) let \(\mathcal{A}_{i}\) be the event that \(a\) is divisible by \(p_{i},\) for \(i=1, \ldots, r\) (a) Show that \(\varphi(n) / n=\mathrm{P}\left[\overline{\mathcal{A}}_{1} \cap \cdots \cap \overline{\mathcal{A}}_{r}\right],\) where \(\varphi\) is Euler's phi function. (b) Show that if \(J \subseteq\\{1, \ldots, r\\},\) then $$ \mathrm{P}\left[\bigcap_{j \in J} \mathcal{A}_{j}\right]=1 / \prod_{j \in J} p_{j} $$ Conclude that \(\left\\{\mathcal{A}_{i}\right\\}_{i=1}^{r}\) is mutually independent, and that \(\mathrm{P}\left[\mathcal{A}_{i}\right]=1 / p_{i}\) for each \(i=1, \ldots, r\) (c) Using part (b), deduce that $$ \mathrm{P}\left[\overline{\mathcal{A}}_{1} \cap \cdots \cap \overline{\mathcal{A}}_{r}\right]=\prod_{i=1}^{r}\left(1-1 / p_{i}\right) . $$ (d) Combine parts (a) and (c) to derive the result of Theorem 2.11 that $$ \varphi(n)=n \prod_{i=1}^{r}\left(1-1 / p_{i}\right) $$

Suppose \(X\) and \(X^{\prime}\) are random variables that take values in a set \(S\) and that have essentially the same distribution. Show that if \(f: S \rightarrow T\) is a function, then \(f(X)\) and \(f\left(X^{\prime}\right)\) have essentially the same distribution.

Let \(\mathcal{B}\) be an event, and let \(\left\\{\mathcal{B}_{i}\right\\}_{i \in I}\) be a finite, pairwise disjoint family of events whose union is \(B\). Generalizing the law of total probability (equations (8.9) and (8.10)\(),\) show that for every event \(\mathcal{A},\) we have \(\mathrm{P}[\mathcal{A} \cap \mathcal{B}]=\) \(\sum_{i \in I} \mathrm{P}\left[\mathcal{A} \cap \mathcal{B}_{i}\right],\) and if \(\mathrm{P}[B] \neq 0\) and \(I^{*}:=\left\\{i \in I: \mathrm{P}\left[\mathcal{B}_{i}\right] \neq 0\right\\},\) then $$ \mathrm{P}[\mathcal{A} \mid \mathcal{B}] \mathrm{P}[\mathcal{B}]=\sum_{i \in I^{*}} \mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \mathrm{P}\left[\mathcal{B}_{i}\right] $$ Also show that if \(\mathrm{P}\left[\mathcal{A} \mid \mathcal{B}_{i}\right] \leq \alpha\) for each \(i \in I^{*},\) then \(\mathrm{P}[\mathcal{A} \mid \mathcal{B}] \leq \alpha\)

Suppose that \(\left\\{X_{i}\right\\}_{i \in I}\) is a finite, non-empty, mutually independent family of random variables, where each \(X_{i}\) is uniformly distributed over a finite set \(S\). Suppose that \(\left\\{Y_{i}\right\\}_{i \in I}\) is another finite, non-empty, mutually independent family of random variables, where each \(Y_{i}\) has the same distribution and takes values in the set \(S\). Let \(\alpha\) be the probability that the \(X_{i}\) 's are distinct, and \(\beta\) be the probability that the \(Y_{i}\) 's are distinct. Using the previous exercise, show that \(\beta \leq \alpha\).

For real-valued random variables \(X\) and \(Y\), their covariance is defined as \(\operatorname{Cov}[X, Y]:=E[X Y]-E[X] E[Y] .\) Show that: (a) if \(X, Y,\) and \(Z\) are real-valued random variables, and \(a\) is a real number, then \(\operatorname{Cov}[X+Y, Z]=\operatorname{Cov}[X, Z]+\operatorname{Cov}[Y, Z]\) and \(\operatorname{Cov}[a X, Z]=a \operatorname{Cov}[X, Z]\) (b) if \(\left\\{X_{i}\right\\}_{i \in I}\) is a finite family of real-valued random variables, then $$ \operatorname{Var}\left[\sum_{i \in I} X_{i}\right]=\sum_{i \in I} \operatorname{Var}\left[X_{i}\right]+\sum_{i, j \in I \atop i \neq j} \operatorname{Cov}\left[X_{i}, X_{j}\right] $$
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