ece 109 cumsheet
cumulative cheat sheet
i will turn this into a better cheat sheet for the final, just need to know what doesn't need to be on it
week 1: event
i dont have anything to say lmao
P(E \cup F) = P(E) + P(F)  P(EF)
"inclusionexclusion principle," subtracts double counted intersection
P(E) = \frac{E}{S}
but only if it's
equiprobable!!
\binom{n}{k} = \frac{n!}{k! \, (n  k)!}
G  F = GF^C
does not occur = complement occurred
"not necessarily" is a valid response to "does [event] occur?"
postquiz note: none of this was used
week 2: given
conditional probability
P(E  F) = \frac{P(EF)}{P(F)} = \frac{P(F  E) \, P(E)}{P(F)}
P(EF) = P(E  F) \, P(F) = P(F  E) \, P(E)
\begin{aligned} P(B) &= P(BA) + P(BA^C) = P(B  A) \, P(A) + P(B  A^C) \, P(A^C) ~ \text{splitting into cases} \\ &= P(B  A_1) \, P(A_1) + \dots + P(B  A_n) \, P(A_n) \\ P(B  E) &= P(B  A_1 E) \, P(A_1  E) + \dots + P(B  A_n E) \, P(A_n  E) \end{aligned}
P(A_1 \dots A_n) = P(A_1) \, P(A_2  A_1) \, P(A_3  A_1 A_2) \dots
P(A_n  A_1 \dots A_{n1})
"chain rule"
P(A_i  B) = \dfrac{P(B  A_i) \, P(A_i)}{\displaystyle\sum_{j =
1}^n P(B  A_j) \, P(A_j)}
"bayes theorem"
if E
and
F
disjoint,
P(E  F) = 0
if
E \subseteq F
then
EF = E
,
P(F  E) = 1
split complicated events into cases
post quiz note: we did not use this. did need to know how to interpret gibberish like "ur mom but not my mom"
week 3: independent
If E
and F
are
independent,
P(EF) = P(E) \, P(F)

P(E  F) = P(E)
ifP(F) \neq 0

E
andF^C
,E^C
andF
,E^C
andF^C
are also independent
If A
and B
are
independent given C
,

P(AB  C) = P(A  C) \, P(B  C)
ifP(C) \neq 0

P(A  BC) = P(A  C)

but they aren't independent given
C^C
!!
If E
and F
are
disjoint and their probabilities aren't 0, they're
not independent
If three events are independent to each other ("pairwise independent")
and
P(ABC) = P(A) \, P(B) \, P(C)
(which isn't
necessarily true), they're independent
Probability E
occurs exactly
k
times in
n
independent trials:
\binom{n}{k} (P(E))^k (1  P(E))^{nk}
For disjoint A
and
B
, probability that
A
occurs before
B
in independent trials:
\frac{P(A)}{P(A) + P(B)}
and i guess random vars are fair game too??
"with replacement" = putting it back (→ independent trials)
disjoint and subset events are not independent
don't forget that you know about their complements too
post quiz: P(A  C)
is NOT
\frac{P(A)}{P(C)}
. the conditional
independence equations above were useful
week 4: cumdf
(CDF/pmf definitions moved to a table in week 5)
here are a bunch of formulas that are probably kind of easy to remember/figure out on the spot
P(X > u) = 1  F(u)
P(X < u) = F(u^)
P(X \geq u) = 1  F(u^)

P(X = u) =
"jump atu
" (if continuous it's 0) P(a < X \leq b) = F(b)  F(a)
P(a < X < b) = F(b^)  F(a)

P(a \leq X < b) = F(b^)  F(a^)
P(a \leq X \leq b) = F(b)  F(a^)
cumDF only needs to be right continuous, so
F(u) = F(u^+) \neq F(u^)
when zeger has P([a, b]) = b  a
, split into
cases based on possible values of X
or
endpoints of idk (for example, if X
could be
0 and 1, then your cases would be u < 0
,
0 \leq u < 1
, etc, see w4d2 lecture
example or winter 22 quiz 4.3)
also, P(E^C  F) = 1  P(E  F)
i do not imagine any of this will be needed on the quiz
post quiz: nothing was used. the quiz was easy
week 5: expected
function  stands for  typical function name and formula  graph notes  rules/properties 

CDF 
cumulative distribution function 
F_X(u) = P(X \leq u) 
filled bubbles always above open circles 
never decreases, starts at 0 and ends at 1 
pmf 
probability mass function 
p_X(u) = P(X = u) 
bars with dots on them 
never negative, bars add to 1 
probability density function 
f_X(u) = \frac{dF_X(u)}{du} 
never negative, area under curve is 1 P(a < X \leq b) = \int\limits_a^b f_X(u) du 
expected value/mean:
E[X] = \sum_u u f_X(u)
E[X] = \int\limits^\infty_{\infty} u f_X(u) du
special rv  pmf/pdf  intuition  mean  variance  

uniform discrete rv 


binary rv
(deterministic rv when 
p_X(0) = 1  q p_X(1) = q 
flipping a biased coin 
E[X] = q 
\sigma_X^2 = q(1q) 

binomial rv 
p_X(k) = \binom{n}{k} p^k (1  p)^{nk} for 
number of heads from flipping biased coin

E[X] = np 

poisson rv 
p_X(k) = \dfrac{\lambda^k e^{\lambda}}{k!} for 
E[X] = \lambda 

geometric rv 
p_X(k) = p(1  p)^{k  1} for 
number of biased coin flips until first head 
E[X] = \frac{1}{p} 

uniformly distributed rv on 
f_X(u) = \begin{cases} \dfrac{1}{b  a} & \text{if $a < u < b$} \\ 0 & \text{else} \end{cases} 
E[X] = \frac{a + b}{2} 
\sigma_X^2 = \frac{(ab)^2}{12} 

gaussian/normal rv X \sim \N(m, \sigma^2) 
f_X(u) = \frac{1}{\sigma \sqrt{2\pi}} e^{\dfrac{1}{2}\left(\dfrac{u  m}{\sigma}\right)^2} 
E[X] = m 
\Var(X) = \sigma^2 

exponential rv 
f_X(u) = \begin{cases} \lambda e^{\lambda u} & \text{if $u > 0$} \\ 0 & \text{else} \end{cases} 
E[X] = \frac{1}{\lambda} 
postquiz: none of this was very useful, but desmos is!!
week 6: variance
CDF for X \sim \N(0, 1)
:
\Phi(u) = \frac{1}{\sqrt{2\pi}} \int\limits_{\infty}^u e^{\frac{z^2}{2}} dz = \frac{1}{2} + \frac{1}{2}\erf\left(\frac{u}{\sqrt{2}}\right)
for X \sim \N(m, \sigma^2)
,
F_X(u) = \Phi\left(\dfrac{um}{\sigma}\right)
i dont think i need to know this at all, but
\erf(u) = \frac{2}{\sqrt{\pi}} \int\limits_0^u e^{t^2} dt
more expected value properties:
E[g(X)] = \begin{cases} \displaystyle\int\limits_{\infty}^\infty g(u) f_X(u) du & \text{continuous} \\ \displaystyle\sum_u g(u) p_X(u) & \text{discrete} \end{cases}
E[g_1(X) + g_2(X)] = E[g_1(X)] + E[g_2(X)]
n
th moment:
E[X^n]
. first moment is mean
n
th central moment:
E[(X  E[X])^n]
. first central moment is
zero, second central moment is variance (sqrt is stddev)
\sigma_X^2 = \Var(X) = E[(X  E[X])^2] = E[X^2]  E[X]^2
for Y = aX + b
:
E[aX + b] = aE[X] + b

pmf
p_Y(u) = p_X\left(\dfrac{ub}{a}\right)

variance
\sigma_Y^2 = a^2 \sigma_X^2

if
X \sim \N(m, \sigma^2)
, thenY \sim \N(am + b, a^2 \sigma^2)
to find pdf of Y
in terms of
p_X(u)
(pdf of X
)

set up CDF of
Y
(getF_Y
in terms ofF_X
) differentiate
\frac{d}{du} \int\limits_{h(u)}^{g(u)} f(z, u) dz = f(g(u), u) g'(u)  f(h(u), u) h'(u) + \int\limits_{h(u)}^{g(u)} \left(\frac{\partial}{\partial u} f(z, u)\right) dz ~ \text{(leibniz rule)}
joint cumdfs probably not this week
You can visually use the center of mass of the pdf/pmf for expected value
Variance is never negative
Use Wolfram Alpha!
be prepared to complete the square and stuff to make things look like a gaussian
post quiz note:
know your log properties

\Var(aX^2) = a^2\Var(X^2)
because if you letY = X^2
then\Var(aX^2) = \Var(aY) = a^2 \Var(Y) = a^2 \Var(X^2)

when using desmos to calculate variance of
Y = 2  3X^2
you can't dropX = 0
when calculating expected value becauseY \neq 0
week 7: joint
joint  marginal  

cumDF 
F_{X,Y}(u, v) = P(X \leq u, Y \leq v) 
F_X(u) = F_{X,Y}(u, \infty) F_Y(u) = F_{X,Y}(\infty, v) 
pmf 
p_{X,Y}(u, v) = P(X = u, Y = v) to get a probability of a region, just add the probabilities in it P((X, Y) \in T) = \sum_{(u, v) \in T} p_{X,Y}(u, v)
the CDF is like an infinite rectangle with top right corner
F_{X,Y}(a, b) = P((X, Y) \in \{(u, v)  u \leq a, v \leq b\}) \sum_{u, v} p_{X,Y}(u, v) = 1 
p_X(u) = \sum_u p_{X,Y}(u, v) p_Y(v) = \sum_u p_{X,Y}(u, v) 
you can take a partial derivative in either order f_{X,Y}(u, v) = \frac{\partial^2 F_{X,Y}(u, v)}{\partial u \, \partial v} derived from P((X, Y) \in T) = \iint\limits_T f_{X,Y}(u, v) \, du \, dv pdf ⇒ CDF: F_{X,Y}(u, v) = \int\limits_{\infty}^v \int\limits_{\infty}^u f_{X,Y}(s, t) \, ds \, dt \int\limits_{\infty}^\infty \int\limits_{\infty}^\infty f_{X,Y}(u, v) \, du \, dv = 1 
f_X(u) = \int\limits_{\infty}^\infty f_{X,Y}(u, v) \, dv f_Y(v) = \int\limits_{\infty}^\infty f_{X,Y}(u, v) \, du 
(from spencer vid) to find a probability from joint pdf:

figure out what region to integrate over depending on the event given
do integral. lmao
shortcut! if you know that a marginal pdf will be constant (eg when
finding marginal pdf of Y
where joint pdf
only dependent on u
(X
)) you can just solve for C
by setting area
= 1
post quiz. desmos is ideal for those double integrals. everything was just pdf stuff, so didnt use this cumsheet for the quiz
week 8: independent rvs
these are equivalent:

rvs
X
andY
are independent 
P(X \in A, Y \in B) = P(X \in A) \, P(Y \in B)
where
A, B \in \mathbb{R}

joint CDF factors:
F_{X,Y}(u, v) = F_X(u) \, F_Y(v)
for allu, v

joint pdf factors:
f_{X,Y}(u, v) = f_X(u) \, f_Y(v)
for allu, v

joint pmf factors:
p_{X,Y}(u, v) = p_X(u) \, p_Y(v)
for allu, v
iid (independent, identically distributed) means same pdf/pmf
rv  

Z = X + Y 
\begin{aligned} f_e(w) &= \int\limits_{\infty}^\infty f_{X,Y}(w  v, v) \, dv \\ &= \int\limits_{\infty}^\infty f_{X,Y}(u, w  u) \, du \\ &= \int\limits_{\infty}^\infty f_X(w  v) \, f_Y(v) \, dv ~ \text{(if independent; convolution integral)} \\ &= \int\limits_{\infty}^\infty f_X(u) \, f_Y(w  u) \, du \end{aligned} 
Z = \max(X, Y) 
\begin{aligned} f_Z(w) &= \int\limits_{\infty}^t f_{X,Y}(u, t) \, du + \int\limits_{\infty}^t f_{X,Y}(t, v) \, dv \\ &= F_X(t) \, f_Y(t) + f_X(t) \, F_Y(t) ~ \text{(if independent)} \\ &= n \, F_X^{n1}(t) \, f_{X_1}(t) ~ \text{(if $n$ iid rvs)} \\ &= \begin{cases} nt^{n1} ~ \text{if $t \in [0, 1]$} \\ 0 ~ \text{else} \end{cases} ~ \text{(if rvs uniform on $[0, 1]$)} \end{aligned} 
Z = XY 
f_Z(a) = \int\limits_{\infty}^\infty \frac{1}{v} f_{X,Y}\left(\frac{a}{v}, v\right) \, dv 
from spencer:

P(\max(X, Y) \geq 1) = 1  P(\max(X, Y) < 1)
which is a bit more useful 
polar integration
\iint f(r, \theta) r dr d\theta
when the quiz asks which of a bunch of joint pmf graphs are independent, the rows/columns should be multiples of each other
postquiz: yeah actually in general, if you have two independent uniform rvs, their joint pdf/pmf will be a rectangle/array
week 9: covariance
f_{X + Y, X  Y}(a, b) = \frac{1}{2} f_{X,Y}\left(\frac{a + b}{2}, \frac{a  b}{2}\right)
\begin{aligned} E[Z = g(X_1, \dots, X_n)] &= \int\limits_{\infty}^\infty \int\limits_{\infty}^\infty \dots \int\limits_{\infty}^\infty g(u_1, \dots, u_n) \, f_{X_1, \dots, X_n}(u_1, \dots, u_n) \, du_1 \dots du_n \\ E[g(X, Y)] &= \int\limits_{\infty}^\infty \int\limits_{\infty}^\infty g(u, v) \, f_{X, Y}(u, v) \, du dv & \text{(continuous)} \\ &= \sum_v \sum_u g(u, v) \, p_{X, Y}(u, v) & \text{(discrete)} \end{aligned}
E[X + Y] = E[X] + E[Y] ~ \text{They don't need to be independent}
covariance
\begin{aligned} \Cov(X, Y) &= E[(X  E[X])(Y  E[Y])] \\ &= E[XY]  E[X] \, E[Y] \\ \Cov(X, X) &= \Var(X) \\ \Cov(X, Y + Z) &= \Cov(X, Y) + \Cov(X, Z) \\ \Cov(aX, bY) &= ab \Cov(X, Y) \\ \Var(X \pm Y) &= \Var(X) + \Var(Y) \pm 2 \Cov(X, Y) \\ &= \Var(X) + \Var(Y) ~ \text{if uncorrelated} \end{aligned}
for independent rvs,
\Var(X_1 \pm \dots \pm X_n) = \Var(X_1) \pm \dots \pm \Var(X_n)
correlation coefficient
\begin{aligned} \rho_{X, Y} &= \frac{\Cov(X, Y)}{\sigma_X \sigma_Y} \\ &= E\left[\left(\frac{X  E[X]}{\sigma_X}\right)\left(\frac{Y  E[Y]}{\sigma_Y}\right)\right] \\ \rho_{X, Y} &\leq 1 \end{aligned}
independence → (uncorrelated ↔
E[XY] = E[X] \, E[Y]
)
then zeger went on to talk about estimation.. which spencer DID mention , so ðŸ˜±ðŸ˜±ðŸ˜±
mean squared error = E[(X  \hat{X})^2]
with
error e
, true value
X
, estimate
\hat{X}
best linear estimate
\begin{aligned} \hat{X} &= \frac{E[XY]}{E[Y^2]} \cdot Y \\ &= \rho_{X, Y} Y ~ \text{if $\sigma_X = \sigma_Y = 1$, $E[X] = E[Y] = 0$} \end{aligned}
joint gaussian
\begin{aligned} f_{X,Y}(u, v) &= \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1\rho^2}} ~ e^{\displaystyle \frac{1}{2(1  \rho^2)} \left( \left(\frac{u  m_X}{\sigma_X}\right)^2 + \left(\frac{v  m_Y}{\sigma_Y}\right)^2  2\rho\left(\frac{u  m_X}{\sigma_X}\right)\left(\frac{v  m_Y}{\sigma_Y}\right) \right)} ~ \text{where $\rho$ is the correlation coeff} \\ &= \frac{1}{2\pi\sqrt{1\rho^2}} ~ e^{\displaystyle \frac{1}{2(1  \rho^2)} (u^2 + v^2  2\rho uv)} ~ \text{if $m_X = m_Y = 0$, $\sigma_X = \sigma_Y = 1$} \\ \end{aligned}
just for gaussians, independence ↔ uncorrelated
don't mix up stddev and variance!! make sure to sqrt variance for
\sigma_X
for correlation coefficient, but
when the problem asks to sum variances leave it as
\sigma_X^2
maybe the infinite integral of e^{ax}
might
be nice to have
\begin{aligned} \int\limits_0^\infty e^{ax} dx &= \frac{1}{a} e^{ax} \bigg\rvert_0^\infty \\ &= \frac{1}{a} e^{a\infty} + \frac{1}{a} e^0 \\ &= \frac{1}{a} \cdot 0 + \frac{1}{a} \cdot 1 \\ &= \boxed{\frac{1}{a}} \end{aligned}
actually this seems incredibly limited i would just use desmos
at least for fall '21 #2 apparently it is better to use the gaussian pdf not the joint gaussian
postquiz: needed to double check variance equation