---
# System prepended metadata

title: HW 1

---

# ML HW 1

Ivan Christian

## 1

### a)



| A/B | 1 | 2 | 3 |
| -------- | -------- | -------- | -------- |
| 2     | $\frac{1}{36}$     | $\frac{5}{18}$     |$\frac{1}{9}$
| 3     | $\frac{1}{4}$     | $\frac{5}{36}$     |$\frac{7}{36}$

$P(A=3, B = 3)$ = $\frac{7}{36}$

### b)
$P(B = 1)$ = $\frac{1}{36} + \frac{1}{4}$ = $\frac{5}{18}$
$P(B = 2)$ = $\frac{5}{18} + \frac{5}{36}$= $\frac{5}{12}$
$P(B = 3)$ = $\frac{1}{9} + \frac{7}{36}$ = $\frac{11}{36}$


### c)

$E(X)$ = $\sum{x . P(X)}$ = $1 * \frac{5}{18} + 2 * \frac{5}{12} + 3 * \frac{11}{36}$ = $\frac{73}{36}$ = $2\frac{1}{36}$

---------------
$E(X^2)$ = $\sum{x^2 . P(X)}$ = $1^2 * \frac{5}{18} + 2^2 * \frac{5}{12} + 3^2 * \frac{11}{36}$

$E(X^2)$ = $\frac{5}{18} +  \frac{20}{12} +  \frac{99}{36} = \frac{169}{36}$ 

$Var(X) = E(X^2) - E(X)^2$
$Var(X) = \frac{169}{36} - (\frac{73}{36})^2 = 0.58256172839$

### d)

$P(A = 2 | B = 1) = \frac{P(A=2, B = 1)}{P(B=1)} = \frac{\frac{1}{36}}{\frac{5}{18}} = 0.1$

## 2


### a)

#### Range of the parameters
Max of $P(X=1)= 1$
Min of $P(X=1)= 0$

$0 ≤ θ ≤ \frac{1}{2}$ 

Max of $P(X=2)= 1$
Min of $P(X=2)= 0$

$0 ≤ θ ≤ \frac{1}{3}$ 

Max of $P(X=3)= 1$
Min of $P(X=3)= 0$

$-\frac{1}{5} ≤ θ ≤ \frac{1}{5}$, we know that probability has a minimum of 0 therefore,

$0 ≤ θ ≤ \frac{1}{5}$


#### Maximum Likelihood Estimator

$L(θ) = P(X=1)*P(X=2)*P(X=3)$
$L(θ) = 2θ^{x_1}*3θ^{x_2}*(1-5θ)^{x_3}$
$log(L(θ)) = log(2θ^{x_1}*3θ^{x_2}*(1-5θ)^{x_3})$
$log(L(θ)) = x_1log(2θ) +x_2log(3θ) + x_3log(1-5θ)$


Maximum value is found when the derivative is 0


$\frac{d}{dθ}(log(L(θ))) = \frac{d}{dθ}(x_1log(2θ) +x_2log(3θ) + x_3log(1-5θ)) = 0$

$\frac{x_1 + x_2}{θ} = \frac{5x_3}{1-5θ}$

$θ̂= \frac{x_1+x_2}{5(x_1+x_2+x_3)}$

To find the maximum likelihood estimator:
$\frac{d^2}{dθ}  log⁡(L(θ))= -\frac{x_1}{θ^2} -\frac{x_2}{θ^2} +\frac{(25x_3)}{(1-5θ)^2}$ 

So long as $\frac{d^2}{dθ} log⁡(L(θ)) < 0$ , $θ$ would be the maximum value of the likelihood.


### b)

Let $L(a,b|X_1,…,X_n )=∏_{(i=1)}^{n}\frac{1}{(b-a)}  1_{(X_i∈[a,b])}$

where, $1_{(X_i∈[a,b])}$ is the indicator function, where the value is 1 when $a≤X_i≤b$


$L(a,b|X_1,…,X_n)= (\frac{1}{b-a})^n$
$log(L(a,b|X_1,…,X_n)= log((\frac{1}{b-a})^n)$
$log(L(a,b|X_1,…,X_n)= -nlog(b-a)$

$\frac{d}{da}(log(L(a,b|X_1,…,X_n)) = \frac{n}{b-a} = 0$
$\frac{d}{db}(log(L(a,b|X_1,…,X_n)) = \frac{n}{a-b} = 0$

we know that $n>0$ and therefore the derivative is inconclusive in determining hte maximum likelihood estimator. As such to maximise the likelihood

$a ̂=min⁡ X_i$ and $b ̂=max⁡ X_i$
## 3

### a) 

$μ_1=\frac{(2+8)}{2}=5$
$μ_2=\frac{(1+3+9+10)}{4}=5.75$
### b)
$D_1=\{1,2,3\}$
$D_2=\{8,9,10\}$
### c)

$μ_1=\frac{(1+2+3)}{3}=2$
$μ_2=\frac{(8+9+10)}{3}=9$

Clustering is stable since the new possible centroid are the same. 

$D_1=\{1,2,3\}$
$D_2=\{8,9,10\}$

## 4

### a)

1. To be a probability density function one of the criteria is that  the sum of a probability function should equal to 1, which is the are/volume calculated by integrating within the limits.

$=\int_{0}^{1}\int_{0}^{2} \frac{6}{7} (x^2+ \frac{xy}{2}) \,dy\,dx$
$=\frac{6}{7}\int_{0}^{1}[x^2y +\frac{xy^2}{4} ]_0^2,dx$
$=\frac{6}{7}\int_{0}^{1}2x^2 + x\,dx$
$=\frac{6}{7}[\frac{4}{6}x^3 + \frac{3}{6}x^2]_{0}^{1}$
$= 1$

2. $f(x,y)≥0$ for $x∈[0,1]$ and $y∈[0,2]$ also shows that this pdf is valid

### b)
PDF of X (Marginal probability of X)

$g(x)=∫_0^2f(x,y)dy$
$g(x)=∫_0^2\frac{6}{7} (x^2+ \frac{xy}{2}) dy$
$g(x)=[\frac{6}{7} (x^2y+ \frac{xy^2}{4})]_0^2$
$g(x)=[\frac{6x^2y}{7} + \frac{3xy^2}{14})]_0^2$
$g(x)=[\frac{12x^2}{7} + \frac{6x}{7}] \ 0 ≤ x ≤ 1, = 0 \ otherwise$

### c)

Hence, using the pdf of X

$f_{Y|X} (y | x=1) = \frac{f(x=1,y)}{f(x=1)}$
$f_{Y|X} (y | x=1)=\frac{\int_0^y 6/7(1 + 0.5y) dy}{18/7}$
$f_{Y|X} (y | x=1)=\frac{\int_0^y (1 + 0.5y) dy}{3}$



### d)

To check for independence $f(x,y) = f_X(x)*f_Y(y) = g(x) g(y)$

Marginal pdf Y

$g(y)=∫_0^1f(x,y)dx$
$g(y)=∫_0^2\frac{6}{7} (x^2+ \frac{xy}{2}) dx$
$g(y)=[\frac{6}{7}(\frac{x^3}{3}+\frac{x^2y}{4})]_0^2$
$g(y)=\frac{6}{7}(\frac{8}{3}+y)$

From a glance we can see that $f(x,y) ≠ f_X(x)* f_Y(y)$

Therefore, they are not independent.

## 5


### a)

```
fprintf("Question 5a\n") 
RIASEC = readtable('RIASEC.csv');
only_R = table2array(RIASEC(:,2:9));

m = size(only_R,1);
row_to_remove = [];
for row = 1:m
    if any(only_R(row,:)<0,2) % returns true if row contains any value less than 0
        row_to_remove = [row_to_remove,row];
    end
end
only_R(row_to_remove,:) = [];
```

### b)
```
fprintf("Question 5b\n") 
total_R = sum(only_R,2)/8;
R1 = only_R(:,1);

x = [ones(6500,1) R1(1:6500)];
y = total_R(1:6500);

theta = inv(x.'*x)*x.'*y; % theta0 = 8.1448, theta1 = 3.3888
disp("estimated regression function: " + "y = "+ theta(1) + "+" + theta(2) + "x")

predicted = theta(1) + theta(2)*x(:,2);

RSS = sum((y - predicted).^2);
disp("Residual sum of squares: " + RSS)
RSS_avg = RSS/size(y,1);
disp("Residual sum of squares averaged: " + RSS_avg)
```

![](https://hackmd.io/_uploads/HkH0M4xb6.png)



### c)

```
fprintf("Question 5c\n") 
test_x = R1(6501:end);
test_y = total_R(6501:end);
test_predicted = theta(1) + theta(2)*test_x;
test_RSS = sum((test_y - test_predicted).^2);
disp("Residual sum of squares of remaining people: " + test_RSS)
test_RSS_avg = test_RSS/(size(test_y,1));
disp("Residual sum of squares of remaining people averaged: " + test_RSS_avg)
disp("Residual sum of squares for part b: " + RSS_avg)
if RSS_avg < test_RSS_avg
    disp("RSS of test (part c) is more than training (part b)")
end
```

![](https://hackmd.io/_uploads/H1rkQVeba.png)
