STT 861 Theory of Prob and STT I Lecture Note - 7

2017-10-18

Survival function and its example; transformation of random variables, scale and rate parameters and their examples; Joint probability density and its example; independent continuous random variables.

Portal to all the other notes

Lecture 07 - Oct 18 2017

Survival function

Definition: Survival function $S$ of a r.v. $X$ is 1-CDF.

S (x) = 1 - F (x) = P (X > x)

if $F$ is the CDF of $X$ .

This is meaningful especially if $X$ is a waiting time. Indeed $S (x)$ is the chance of survival to “age” $x$ .

Example 1

Memoryless property of the exponential distribution. Let $X \sim E x p (λ)$ . We know the CDF $F (x) = 1 - e^{- λ x}$ . Then

S (x) = 1 - F (x) = e^{- λ x}

Important: this characterizes the exponential distribution.

Q: What is $P (X + h | X > x)$ (P(Given X survival to age $x$ , what is the chance of survival $h$ units of time longer))?

\begin{aligned} P & = \frac{P ({X > x} \cap {X > x + h})}{P (X > x)} \\ = \frac{P (X > x + h)}{P (X > x)} \\ = \frac{S (x + h)}{S (x)} \\ = e^{- λ h} \end{aligned}

This proves that for $X \sim e x p (λ)$ , the information of how long we have survived tells us nothing about the likelihood of surviving $h$ units of time longer (Nothing to do with $x$ , the current “age”).

FACT: in a Poisson( $λ$ ) process, the waiting time between two jumpers (arrivals) are i.i.d. $E x p (λ)$ r.v.’s.

Sketch of proof: Let’s prove first that $X_{1} \sim E x p (λ)$ .

This $X_{1}$ is the first time that $N$ reaches 1.

The event ${X_{1} < t}$ is the same as $N_{t} \geq 1$ .

Therefore,

\begin{aligned} P (X_{1} < t) & = P (N_{t} \geq 1) \\ = 1 - P (N_{t} = 0) \\ = 1 - e^{λ t} \end{aligned}

We also see this is the cdf of $X_{1}$ . We recognize the formula as $F_{X_{1}} (t) = 1 - e^{- λ t}$ .

Now, let’s convince ourselves that $X_{2} \sim E x p (λ)$ and $X_{1}$ and $X_{2}$ are independent. This is true because $N$ has independent increments. This is related to the memoryless property of $X_{i}$ ’s.

Example of transformation of r.v.’s

Let $X \sim E x p (1)$ , let $λ > 0$ be fixed (not random). Let $Y = \frac{1}{λ} X$ . Let’s find the distribution of $Y$ via its CDF.

S_{Y} (y) = P (Y > y) = P (\frac{x}{λ}) = P (x > λ y)

= {\begin{matrix} e^{- λ y} & y \geq 0 \\ 1 & y < 0 \end{matrix}

We recognize, via this survival function, that $Y$ has the same CDF as $E x p (λ)$ .

Example 2

Let $U \sim U n i f o r m (0, 1)$ . Let $Y = - \ln U$ . Find $Y$ ’s law (distribution) via the survival function. (Note: when $\ln U < 0$ , $Y > 0$ )

\begin{aligned} P (Y > y) & = P (- \ln U > y) \\ = P (\frac{1}{U} > e^{y}) \\ = P (U < e^{- y}) \\ = e^{- y} (F(U)=u) \end{aligned}

(Because the exp function is strictly increasing)

We recognize that $Y \sim E x p (1)$ .

Note: if we have $U \sim U n i f o r m (0, 1)$ , and we want to create a r.v. $Z \sim E x p (λ)$ , we can take $Z = - \frac{1}{λ} \ln (U)$ .

Example of scale and rate parameters

We saw that if we multiply an $E x p$ r.v., with parameter $λ$ by a constant $c$ , we get $E x p$ with parameter $\frac{λ}{c}$ .

For instance,

Y \sim E x p (λ), Y = \frac{1}{λ X}

Z = c Y = \frac{c}{λ X} \sim E x p (\frac{λ}{c})

This shows that $\frac{1}{λ}$ is a scale parameter.

We also say $λ$ is a rate parameter when $X$ is multiplied by $c$ .

For instance, $λ$ is the rate of arrivals of the Poisson r.v. (or process) in a time interval of length 1. But $\frac{1}{λ}$ is the average waiting time for the next arrival. It is the scale of the waiting time.

Generally speaking, a r.v. $X$ (or its law) has a scale parameter $α$ if the CDF of $X$ has this form

F_{X} (x) = \bar{F} (\frac{x}{α})

We also say $X$ has a rate parameter $λ$ if the CDF of $X$ has this form

F_{X} (x) = \tilde{F} (λ x)

Example 3

We just saw, for $X \sim E x p (λ)$ ,

F_{X} (x) = 1 - e^{- λ x}

\tilde{F} (y) = 1 - e^{- y}

Also

F_{X} (x) = 1 - e^{- \frac{x}{1 / λ}}

\bar{F} (y) = 1 - e^{- y}

and $\frac{1}{λ}$ is the scale paremeter $α$ .

Theorems: if $α$ is a scale parameter for $X$ , then

f_{X} (x) = \frac{1}{α} f (\frac{x}{α})

( $α = 1$ is the special case)

If $λ$ is a rate parameter for $X$ , then

f_{X} (x) = λ f (λ x)

( $λ = 1$ is the special case)

Then, let $X_{α}$ have scale parameter $α$ , then

E (X_{α}) = α E (X_{1})

$X_{1}$ is $X_{α}$ with $α = 1$ .

V a r (X_{α}) = α^{2} V a r (X_{1})

let $X_{λ}$ have rate parameter $λ$ , then

E (X_{λ}) = \frac{1}{λ} E (X_{1})

V a r (X_{λ}) = \frac{1}{λ^{2}} V a r (X_{1})

Theorem: Arbitrary function of random variables.

Let $X$ have density $f_{X}$ . Let $Ψ$ be a strictly monotone function (increasing or decreasing). So $Ψ$ ’ the derivative basically exists, and the inverse function $Ψ^{- 1}$ exists. Let $Y = Ψ (X)$ . Then $Y$ has this density.

f_{Y} (y) = f_{X} (Ψ^{- 1} (y)) \frac{1}{Ψ^{'} (Ψ^{- 1} (y))}

Idea of proof:

Start from CDF, of $Y$ and use chain rule.

Exercise: Use this theorem to check the $- \ln U \sim E x p (1)$ when $U \sim U n i f o r m (0, 1)$ .

Definition: (Joint probability density) A pair of r.v. $(X, Y)$ has a joint probability density $f_{X, Y} (x, y)$ (a function on $R^{2}$ ) if

P (X \in [a, b] a n d Y \in [c, d]) = \int_{c}^{d} (\int_{a}^{b} f_{X, Y} (x, y) d x) d y

Definition (More like a theorem): if $f_{X, Y} (x, y)$ as above is really $= f_{X} (x) f_{Y} (y)$ , then $X$ and $Y$ are independent.

Proof:

\begin{aligned} P (X \in d x \cap Y \in d y) & = f_{X, Y} (x, y) d x d y \\ = f_{X} (x) f_{Y} (y) d x d y \\ = f_{X} (x) d x f_{Y} (y) d y \\ = P (X \in d x) P (Y \in d y) \end{aligned}

It also proves that $f_{x}$ is the density of $X$ and $f_{Y}$ is the density of $Y$ .

Example 4

Let $f_{X, Y} (x, y) = 15 e^{- 5 x - 3 y}$ if $x \geq 0, y \geq 0$ , and 0 otherwise. We see

f_{X, Y} (x, y) = 5 e^{- 5 x} \times 3 e^{- 3 y} = f_{X} (x) \times f_{Y} (y)

(the places where $f_{X}$ , $f_{Y}$ and $f_{X, Y} = 0$ , do match up) See $X$ and $Y$ independent and $X \sim E x p (5)$ and $Y \sim E x p (3)$ .

Example 5

Let $f_{X, Y} (x, y) = 1$ when $x \in [0, 1]$ and $y \in [0, 1]$ , =0 otherwise. Then $f_{X, Y} (x, y) = f_{X} (x) \times f_{Y} (y)$ , each of them is i.i.d $U n i f o r m (0, 1)$ .

Example 6

Let $f_{X, Y} (x, y) = c o n s t$ if $0 \leq x \leq y \leq 1$ , =0 otherwise. Then the area in a 2D surface is a triangle. The constant is 2. We say that $(X, Y)$ is uniform in that triangle. However, $X$ and $Y$ are not independent. Because we have

X \leq Y

This is a non-random relation, so $X$ and $Y$ are dependent.

General theorem: For most shapes in $R^{2}$ we can define the uniform density on that shape if its area is finite, with parameter of $\frac{1}{a r e a}$ . Moreover, if the shape has a center of symmetric then $X$ and $Y$ are independent. When the shape is the rectangle $[a, b] \times [c, d]$ . Then $X$ independent $Y$ and $X \sim U n i f o r m [a, b]$ , $Y \sim [c, d]$

For a circle shape, $X$ and $Y$ are not independent, but $ρ$ and $θ$ are independent.

X_{1} = ρ \cos θ, Y_{1} = ρ \sin θ, (ρ, θ)

are independent.

θ \sim U n i f o r m [0, 2 π]

Exercise: find the density of $ρ$ .

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