```
dice = 1:6
using Statistics
μ = Statistics.mean(dice)
sum(dice) / length(dice)
Statistics.var(dice, corrected=false)
sum((dice .- μ).^2 / length(dice))
Statistics.var(dice)
Statistics.std(dice)^2
sum((dice .- μ).^2 / (length(dice) - 1))
[1:6; 1:6]
Statistics.cor([1:6; 1:6])
[1:6; 6:-1:1]
Statistics.cor([1:6; 6:-1:1])
[1:6 6:-1:1]
Statistics.cor([1:6 6:-1:1])
```

`rand`

produces independent output so we should see a zero correlation. For small values we still see a bit of correlation

```
rand(6, 2)
Statistics.cor(rand(6, 2))
```

But as we increase, the correlation goes to zero

```
Statistics.cor(rand(100, 2))
Statistics.cor(rand(100000, 2))
```

```
using Plots
μ = 0
σ = 1
f(x) = exp(-1/2*((x - μ) / σ)^2) / (σ * √(2π))
x = range(-4.0, stop=4.0, length=1000)
plot(x, f.(x))
z = randn(100)
scatter!(z, f.(z), markersize=2)
```

*See Example 7.1 of Boyd & Vandenberghe "Convex Optimization" book.*

Suppose we observe $y_i = a_i^\top x + v_i$ where $v_i$ has a normal distribution $\mathcal{N}(\mu_i, \sigma_i^2)$.

We want to recover $x$ from the observation $y_i$ such that the $v_i$ are likely to occur. That is, we want to maximize $f(v_i)$. This is multi-objective. Instead, we can maximize the likelihood that they all occur. Since they are independent, it's the product.

$\prod_{i=1}^n f(v_i) = \prod_{i=1}^n \frac{1}{\sigma_i \sqrt{2\pi}}\exp(-\frac{(v_i - \mu_i)^2}{2\sigma_i^2})$Since $\log$ is an increasing function, that is equivalent to maximizing the logarithm:

$\sum_{i=1}^n \log(\frac{1}{\sigma_i \sqrt{2\pi}}\exp(-\frac{(v_i - \mu_i)^2}{2\sigma_i^2}))$which is equal to

$-\sum_{i=1}^n \log(\sigma_i) - \frac{n\log(2\pi)}{2} -\frac{1}{2\sigma_i^2} \sum_{i=1}^n (v_i - \mu_i)^2$The first two terms do not depend on $v_i$ so we can drop them.

$-\frac{1}{2} \sum_{i=1}^n \frac{(v_i - \mu_i)^2}{\sigma_i^2}$$-\frac{1}{2}$ is a negative constant so **maximizing** this expression is equivalent to **minimizing**

In terms of $x$, this is

$\min_x \sum_{i=1}^n \frac{(y_i - a_i^\top x - \mu_i)^2}{\sigma_i^2}$If $\mu_i = 0$ and $\sigma_i$ does not depend on $i$ (same for all samples, we say they are independent and identially distributed (**i.i.d**)), this gives

where the $i$th row of $A$ is $a_i$. So the classical linear regression we saw during the first week assumes i.i.d. normal noise of zero mean.

How to interpret the scaling $\sigma_i^2$ in terms of influence on $x$ for very noisy samples ?

```
n = 101
σ = [1000; ones(n - 1)]
μ = 100rand(n)
v = randn(n) .* σ .+ μ
m = 10
x = rand(m)
A = rand(n, m)
y = A * x + v
```

Without taking the noise into account, we get large errors:

`x - A \ y`

Taking $\mu$ into account can be done as follows:

`x - A \ (y - μ)`

How do we take $\sigma$ into account ?

We know that `\`

solves the least square

but we want it to solve

$\sum_{i = 1}^n (y_i - a_i^\top x - \mu_i)^2 / \sigma_i^2$instead. How do we do this ? The expression is equal to

$\sum_{i = 1}^n (\frac{y_i - \mu_i}{\sigma_i} - \frac{a_i^\top}{\sigma_i} x)^2$so we can just scale $y$ and $A$:

`x - (A ./ σ) \ ((y - μ) ./ σ)`

Now that's much better.

*This page was generated using Literate.jl.*

© Benoît Legat. Last modified: September 10, 2024. Website built with Franklin.jl.