The cumulative distribution function (CDF) of an random variable $X $ is denoted $F_X $ and is defined as:
$$F_X(x)=\b{P}(X \leq X) = \b{P}(\set{ \omega \in \Omega \mid X(\omega) \leq x})$$
Let $F $ be a cumulative distribution function (CDF), $f $ be the corresponding probability distribution function (PDF).
- $F $ is non-decreasing.
- $\limu{x}{ -\infty } F(x) = 0$.
- $\limu{x}{ +\infty } F(x) =1$.
- If $X$ is discrete, then $X$ is piecewise constant and continuous from the left.
- If $X$ is continuous and $f_X$ is continuous, then $F$ is differentiable.
- If $X$ is continuous, $F'_X(x) = f_X(x)$.
- if $X$ is discrete and $X$ is integer-valued, then
$$\begin{align*} \b{P}(X=n) &= \b{P}(X \leq n) - \b{P}(X \leq n-1) \br &= F_X(n) - F_X(n-1) \end{align*}$$
$$f_X(x) = \frac{1}{\sqrt{2\pi}{ \sigma }}e^{-\frac{(x-{ \mu })^2}{2{ \sigma }^2}}$$
For any $B \subseteq R $, 2 e.g. condition on value of $X$.
$ f_{X \mid \set{ X \in E }}(x) = \begin{cases} \frac{ f_X(x)}{ \b{P}(x \in E)}, \if x \in E \br 0, \otherwise \end{cases} $
The time $T$ until a call comes into a call center is modeled by an exponential random variable with parameter $\lambda > 0$, you turn on phone, leave for $t$ units of time and return. The phone didn’t ring that time. Let $X$ be additional time until phone rings. Let’s find CDF of $X$ given $T>t$,
$$f_T(s) = \begin{cases} e^{- \lambda s}, s \geq 0 \br 0 , \otherwise \end{cases} $$
$$f_T(s) = \begin{cases} 1-e^{- \lambda s}, s \geq 0 \br 0 , \otherwise \end{cases} $$
We seek $F_{X| T > t} (x) = \b{P}(X \leq x | T > t)$
Note: $t + X = T = \b{P}(T - t \leq x | T > t) $
$$\begin{align*} F_{X| T > t} (x) &= \b{P}(X \leq x | T > t) \br &= 1- \b{P}(X > x | T > t) \br &= 1 - \b{P}(T > x + t | T > t) \br &= 1 - \b{P}(T > x + t | T > t) \br &= 1 - \frac{ \b{P}(T > x + t \cap T > t)}{ \b{P}(T > t)} \br &= 1 - \frac{ P (T > x + t)}{ \b{P}(T > t)} \br &= 1 - \frac{ e^{- \lambda (x + t)}}{ e^{- \lambda t}} \br &= 1 - e^{- \lambda x } \end{align*}$$
This is the CDF of an exponential with parameter $\lambda$ !
Note: doesn’t depend on $t$. This property is special to exponential.
This property is special to exponential, and is called “memoryless”.
Recall: By definition, $E[X | A] = \int_{ -\infty }^{ +\infty } x f_X|A(x) \d x $
Let $A_1, … , A_n$ be a partition of sample space. Then,
1.$f_X(x) = \sum_{ i=1 }^{ n } f_{X \mid A_i}(x) \b{P}(A_i)$
Your arrival time at train stop is uniformly distributed between 7:10 to 7:30am. Trains arrive at 7:15am and 7:30am. Let $y $ be amount of time you wait for train. Let’s find PDF of $Y $ and $E[Y]$.
Let $A $ be the event that you arrive between 7:10 & 7:15am.
$f_Y(y) = f_{y \mid A}(y)\b{P}(A) + f_{y \mid A^c}(y)\b{P}(A^c)$
$f_\set{Y|A}$ is uniform in $[0, 5]$.
$f_\set{Y|A^c}$ is uniform in $[0, 15]$.
so $f_{y|A}(Y) = \begin{cases} \frac{ 1 }{ 5 }, \if y \in[0, 5] \br 0, \otherwise \end{cases}$
so $f_{y|A^c}(Y) = \begin{cases} \frac{ 1 }{ 15 }, \if y \in[0, 15] \br 0, \otherwise \end{cases}$
$\b{P}(A) = \frac{ 5 }{ 20 } = \frac{ 1 }{ 4 }$
For, $y \in [0, 5]$,
$f_Y(y) = \frac{ 1 }{ 5 } \cdot \frac{ 1 }{ 4 } + \frac{ 1 }{ 15 } \frac{ 3 }{ 4 } = \frac{ 1 }{ 10 }$
For, $y \in (5, 15]$,
$f_Y(y) = 0 + \frac{ 1 }{ 15 } \frac{ 3 }{ 4 } = \frac{ 1 }{ 20 }$
so
$$ f_Y(y) = \begin{cases} \frac{ 1 }{ 10 }, y \in [ 0, 5 ] \br \frac{ 1 }{ 20 }, y \in [5, 15] 0, \otherwise \end{cases} $$
Let $A_1, …, A_n$ partitions sample space. Then,
$E[X] \sum_{ i=1 }^{ n } E[X|A_i] \b{P}(A)$
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$$\begin{align*} F[y] &= E[Y|A]\b{P}(A) + E[Y|A^c]\b{P}(A^c) \br &= \frac{ 5 }{ 2 } \frac{ 1 }{ 4 } + \frac{ 15 }{ 2 } \frac{ 3 }{ 4 } \br &= \frac{ 50 }{ 8 } \br &= \frac{ 25 }{ 4 } \end{align*}$$
Let $X, Y $ be continuous random variables. Want to condition $X $ on $Y = y$(say) that $\b{P}(Y = y) = 0$ for any $y$. How to do this?
For any $y$ with $f_Y(y) > 0$. The conditional PDF of $X$ given $Y=y$ is
$$f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{ f_Y(y)}$$
To justify this, can consider for $\delta \approx 0 $
$$\b{P}(x \leq X \leq x + \delta \mid y \leq Y \leq y+ \delta) $$
Apply old definition, and take $\delta \to 0$.
$f_X(x) = \int_{ -\infty }^{ +\infty } f_{X \mid Y}(x|y)f_Y(y) \d y$
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- In examples or modeling, it can be natural to define one random variable and then describe a second one in terms of the first.
i.e. know $f_X, f_{y|X}$.
Can use this to find $f_{X,Y}, f_y, etc. $
e.g. The speed of a typical vehicle driving put a police officer is modeled by an exponential random variable $X$ with mean $50$ mph.
The radar’s measurement of the speed $y$ has an error that’s normal with mean $0 $ and speed equal to $\frac{ 1 }{ 10 }$ of vehicle’s speed. Find $f_{X, Y}$(From this case find $f_Y$)
We see if speed is $x $, then $y = x + N$
note:
$x+N $ is also normal, with mean $x$ and variance (= speed^2)
$$f_{Y \mid X}(y \mid x) = \frac{ 1 }{ \sqrt{2 \pi} \frac{ x }{ 10 }} e^{ \frac{ -(y-x)^2 }{ x^2 /50 }}$$
Since $X $ exponential means,
$$ f_X(x) = \begin{cases} \lambda e^{- \lambda x}a, \if x \geq o \br 0, \otherwise \end{cases} $$
In our situation, $E[X] = 50$, hence, $\frac{ 1 }{ \lambda } = 50$ and $\lambda = \frac{ 1 }{ 50 }$
so $f_X(x) = \begin{cases} \frac{ 1 }{ 50 }e^{-x /50}, \if x \geq 0 \br 0, \otherwise \end{cases}
f_{X, Y}(x,y) = \frac{ 1 }{ sqrt{2 \pi} \frac{ x }{ 10 }} e^{-50 (y-x)^2 / x^2} \frac{ 1 }{ 50 } $