## Contents

- 1 ECE302 Cheat Sheet number 3
- 1.1 Covariance
- 1.2 Correlation Coefficient
- 1.3 Markov Inequality
- 1.4 Chebyshev Inequality
- 1.5 Weak Law of Large Numbers
- 1.6 ML Estimation Rule
- 1.7 MAP Estimation Rule
- 1.8 Bias of an Estimator, and Unbiased estimators
- 1.9 Confidence Intervals, and how to get them via Chebyshev
- 1.10 Definition of the term Unbiased estimators
- 1.11 Few equations from previous material

# ECE302 Cheat Sheet number 3

## Covariance

- $ COV(X,Y)=E[(X-E[X])(Y-E[Y])]\! $
- $ COV(X,Y)=E[XY]-E[X]E[Y]\! $

X and Y are uncorrelated if cov(X,Y) = 0

If X and Y are independent, they are uncorrelated. The converse is not always true.

If either RV X or Y is multiplied by a large constant, then the magnitude of the covariance will also increase. A measure of correlation in an absolute scale is the correlation coefficient.

## Correlation Coefficient

$ \rho(X,Y)= \frac {cov(X,Y)}{\sqrt{var(X)} \sqrt{var(Y)}} \, $ $ = \frac{E[XY]-E[X]E[Y]}{\sqrt{var(X)} \sqrt{var(Y)}} $

-1 <= $ \rho(X,Y) <= 1 $

If $ \rho(X,Y) = 0 $ then x, y are uncorrelated

If $ \rho(X,Y) = 1 or -1 $ then x,y are very correlated

## Markov Inequality

Loosely speaking: In a nonnegative RV has a small mean, then the probability that it takes a large value must also be small.

- $ P(X \geq a) \leq E[X]/a\! $

for all a > 0

EXAMPLE:

On average it takes 1 hour to catch a fish. What is (an upper bound) the probability it will take 3 hours?

SOLUTION:

Using Markov's inequality, where $ E[X]\! $ = 1 and a = 3.

$ P(X \geq 3) \leq \frac {E[X]}{3} = \frac{1}{3} $

so 1/3 is the upper bound to the probability that it will take more than 3 hours to catch a fish.

## Chebyshev Inequality

"Any RV is likely to be close to its mean"

- $ \Pr(\left|X-E[X]\right|\geq C)\leq\frac{var(X)}{C^2}. $

## Weak Law of Large Numbers

The weak law of large numbers states that the sample average converges in probability towards the expected value

- $ \overline{X}_n \, \xrightarrow{P} \, \mu \qquad\textrm{for}\qquad n \to \infty. $

Mn = (X1 + ... + Xn)/n = X1/n + ... + Xn/n

E[Mn] = nE[X]/n = E[X]

Var[Mn] = Var(X1/n) + ... + Var(Xn/n) = Var(X)/n

Pr[ |Mn - E[X]| ] >= Var(Mn)/$ \sigma^2 = Var(X)/n\sigma^2 $

## ML Estimation Rule

$ \hat a_{ML} = \text{max}_a ( f_{X}(x_i;a)) $ continuous

$ \hat a_{ML} = \text{max}_a ( Pr(x_i;a)) $ discrete

If X is a binomial (n,p), where is X is number of heads n tosses,
Then, for any fixed k-value;

$ \hat p_{ML}(k) = k/n $

If X is exponential then it's ML estimate is:

$ \frac{1}{ \overline{X}} $

## MAP Estimation Rule

$ \hat \theta_{MAP} = \text{argmax}_\theta ( f_{\theta|X}(\theta|x)) $

Which can be expanded and turned into the following (if I am not mistaken):

$ \hat \theta_{MAP} = \text{argmax}_\theta ( f_{X|\theta}(x|\theta)f_{\theta}(\theta)) $

for discrete case:

$ \hat \theta_{MAP} = \text{argmax}_\theta ( P_{X|\theta}(x|\theta)P_{\theta}(\theta)) $

## Bias of an Estimator, and Unbiased estimators

An estimator is unbiased if: $ E[\hat a_{ML}] = a $ for all values of a

**A biased estimator can be made unbiased ** : Let an event X be uniform from (0,a). Now $ E[\hat a_{ML}] = a $ = E[X] = a/2,which makes it biased as it is not equal to a. Now we can make $ E[\hat a_{ML}]= a $ if a^{ML}=2x, which makes it unbiased.

## Confidence Intervals, and how to get them via Chebyshev

$ \theta \text{ is unknown and fixed} $

$ \hat \theta \text{ is random and should be close to } \theta \text{ most of the time} $

$ if Pr[|\hat \theta \text{-} \theta|] <= (1-a) \text { then we say we have (1-a) confidence in the interval } [\hat \theta - E, \hat \theta + E] $

Confidence level of $ (1-a) $ if $ Pr[\hat \theta \text{-} \delta < \theta < \hat \theta + \delta] >= (1-a) for all \theta $

## Definition of the term Unbiased estimators

The ML estimator is said to be UNBIASED if its expected value is the true value for all true values.

(ie, A estimate $ \hat \theta $ for parameter $ \theta $ is said to be unbiased if $ \forall \theta \text{(where }\theta\text{ exists)}, E(\hat \theta) = \theta $)

## Few equations from previous material

- X and Y are independent

- $ E[X] = \int^\infty_{-\infty}x*f_X(x)dx\! $

- $ E[XY] = E[X]E[Y]\! $

- $ Var(X) = E[X^2] - (E[X])^2\! $

Marginal PDF

- $ E(g(x))=\int^\infty_{-\infty} g(x)f_X(x,y) dy $

- $ f_X(x) = \int^\infty_{-\infty} f_{XY}(x,y) dy $

- $ f_Y(y) = \int^\infty_{-\infty} f_{XY}(x,y) dx $