Chapter 3 (General Random Variables): Cumulative Distribution Functions (累积分布函数)

本文为 $Introduction$ $to$ $Probability$ 的读书笔记

Cumulative Distribution Functions

• The CDF of a random variable $X$ is denoted by $F_X$ and provides the probability $P(X\le x)$. In particular, for every $x$ we have
  

In what follows, any ambiguous specification of the probabilities of all events of the form { X ≤ x } \{ X \leq x\} { X≤x}, be it through a PMF, PDF, or CDF. will be referred to as the probability law of the random variable $X$.

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Properties of a CDF

• $F_X(x)$ tends to 0 as $x\to -\infty$, and to 1 as $x\to \infty$.
• If $X$ is discrete, then $F_X(x)$ is a piecewise constant function of $x$.
  • If $X$ is discrete and takes integer values, the PMF and the CDF can be obtained from each other by summing or differencing:
    $$\begin{gathered} F_X(k)=\sum _{i=-\infty }^k{p}_X(i) \\ {p}_X(k)=P(X\le k)-P(X\le k-1)=F_X(k)-F_X(k-1) \end{gathered}$$for all integers $k$.
• If $X$ is continuous, then $F_X(x)$ is a continuous function of $x$.
  • The PDF and the CDF can be obtained from each other by integration or differentiation:
    $$\begin{gathered} F_X(x)={\int }_{-\infty }^x{f}_X(t)dt \\ {f}_X(x)=\frac{d{F}_X(x)}{dx} \end{gathered}$$

Sometimes, in order to calculate the PMF or PDF of a discrete or continuous random variable. respectively. it is more convenient to first calculate the CDF.

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Example 3.6. The Maximum of Several Random Variables.
You are allowed to take a certain test three times. and your final score will be the maximum of the test scores. Thus,
X = m a x { X 1 , X 2 , X 3 } , X = max\{X_1, X_2, X_3\}, X=max{ X1​,X2​,X3​},where $X_1,X_2,X_3$ are the three test scores and $X$ is the final score. Assume that your score in each test takes one of the values from 1 to 10 with equal probability $1/10$, independently of the scores in other tests. What is the PMF $p_X$ of the final score?

SOLUTION

• We have
  $$\begin{array}{rl}{F}_X(k)& =P(X\le k)\\ & =P(X_1\le k,X_2\le k,X_3\le k)\\ & =P(X_1\le k)P(X_2\le k)P(X_3\le k)\\ & =(\frac{k}{10}{)}^3\end{array}$$where the third equality follows from the independence of the events { X 1 ≤ k } \{ X_1\leq k\} { X1​≤k}, { X 2 ≤ k } \{X_2\leq k\} { X2​≤k}, { X 3 ≤ k } \{X_3\leq k\} { X3​≤k}. Thus, the PMF is given by
  $$p_X(k)=F_X(k)-F_X(k-1)=(\frac{k}{10}{)}^3-(\frac{k-1}{10}{)}^3,k=1,2,...,10$$
• The preceding line of argument can be generalized to any number of random variables $X_1.....X_n$.
  $$F_X(x)=F_{X_1(x)}...F_{X_n(x)}$$

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Problem 7.
Alvin throws darts at a circular target of radius $r$ and is equally likely to hit any point in the target. Let $X$ be the distance of Alvin’s hit from the center. The target has an inner circle of radius $t$. If $X\le t$, Alvin gets a score of $S=1/X$. Otherwise his score is $S=0$. Find the CDF of $S$. Is $S$ a continuous random variable?

SOLUTION

• For $s<0$, $F_S(s)=0$
• For $0\le s<1/t$, we have
  $$F_S(s)=P(S\le s)=1-P(X\le t)=1-\frac{t^2}{r^2}$$
• For $1/t<s$, the CDF of $S$ is given by
  $$\begin{array}{rl}{F}_S(s)& =P(S\le s)=P(X\le t)P(S\le s|X\le t)+P(X>t)P(S\le s|X>t)\\ & =\frac{t^2}{r^2}P(1/X\le s|X\le t)+(1-\frac{t^2}{r^2})\cdot 1\\ & =\frac{t^2}{r^2}\frac{P(1/s\le X\le t)}{P(X\le t)}+(1-\frac{t^2}{r^2})\\ & =\frac{t^2}{r^2}(1-\frac{1}{s^2{t}^2})+(1-\frac{t^2}{r^2})\\ & =1-\frac{1}{s^2{r}^2}\end{array}$$

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Problem 8.
Consider two continuous random variables $Y$ and $Z$, and a random variable $X$ that is equal to $Y$ with probability $p$ and to $Z$ with probability $1-p$.

• (a) Show that the PDF of $X$ is given by
  $$f_X(x)=p{f}_Y(x)+(1-p)f_Z(x)$$

Note that $E[X]=pE[Y]+(1-p)E[Z]$

• (b) Calculate the CDF of the two-sided exponential random variable that has PDF given by
  
  where $\lambda >0$ and $0<p<1$

SOLUTION

• (a) By the total probability theorem, we have
  $$F_X(x)=P(X\le x)=pP(Y\le x)+(1-p)P(Z\le x)=p{F}_Y(x)+(1-p)F_Z(x)$$By differentiating, we obtain
  $$f_X(x)=p{f}_Y(x)+(1-p)f_Z(x)$$
• (b) Consider the random variable $Y$ that has PDF
  and the random variable $Z$ that has PDF
  We note that the random variables $-Y$ and $Z$ are exponential. Using the CDF of the exponential random variable, we see that the CDFs of $Y$ and $Z$ are given by
  We have $f_X(x)=p{f}_Y(x)+(1-p)f_Z(x)$, and consequently $F_X(x)=p{F}_Y(x)+(1-p)F_Z(x)$. It follows that
  

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Problem 10. Simulating a continuous random variable.
A computer has a subroutine that can generate values of a random variable $U$ that is uniformly distributed in the interval $[0,1]$. Such a subroutine can be used to generate values of a continuous random variable with given CDF $F(x)$ as follows. If $U$ takes a value $u$, we let the value of $X$ be a number $x$ that satisfies $F(x)=u$. For simplicity, we assume that the given CDF is strictly increasing over the range $S$ of values of interest. where S = { x ∣ 0 < F ( x ) < 1 } S = \{ x|0 <F(x) < 1\} S={ x∣0<F(x)<1}. This condition guarantees that for any $u\in (0,1)$, there is a unique $x$ that satisfies $F(x)=u$.

• (a) Show that the CDF of the random variable $X$ thus generated is indeed equal to the given CDF.
  (b) How can this procedure be generalized to simulate a discrete integer-valued random variable?

SOLUTION

• (a) By definition, the random variables $X$ and $U$ satisfy the relation $F(X)=U$. Since $F$ is strictly increasing, we have for every $x$, $X\le x$ if and only if $F(X)\le F(x)$. Therefore,
  $$P(X\le x)=P(F(X)\le F(x))=P(U\le F(x))=F(x)$$
• (b) Let $F$ be the desired CDF. To any $u\in (0,1)$, there corresponds a unique integer $x_u$ such that $F(x_u-1)<u\le F(x_u)$. This correspondence defines a random variable $X$ as a function of the random variable $U$. We then have, for every integer $k$
  $$P(X=k)=P(F(k-1)<U\le F(k))=F(k)-F(k-1)$$Therefore, the CDF of $X$ is equal to $F$, as desired.

The Geometric and Exponential CDFs

• Because the CDF is defined for any type of random variable, it provides a convenient means for exploring the relations between continuous and discrete random variables. A particularly interesting case in point is the relation between geo­metric and exponential random variables.

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• Let $X$ be a geometric random variable with parameter $p$; Thus, for $k=1,2,...,$ we have $P(X=k)=p(1-p{)}^{k-1}$ and the CDF is given by
  $$F_{geo}(n)=\sum _{k=1}^{n}p(1-p{)}^{k-1}=1-(1-p{)}^n,forn=1,2,...$$
• Suppose now that $X$ is an exponential random variable with parameter $\lambda >0$. Its CDF is given by
  $$F_{exp}(x)=P(X\le x)=0,forx\le 0$$and
  $$F_{exp}(x)={\int }_0^x\lambda e^{-\lambda t}dt=1-e^{-\lambda x},forx>0$$
• To compare the two CDFs above, let us define $\delta =-ln(1-p)/\lambda$，so that
  $$e^{-\lambda \delta }=1-p$$Then, we see that the values of the exponential and the geometric CDFs are equal whenever $x=n\delta$, with $n=1,2,....$
  $$F_{exp}(n\delta )=F_{geo}(n),n=1,2,...$$ and are close to each other for other values of $x$.