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Okay.
So, let's look at our matrix equation once

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more.
We have x which is simply draw is the

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data, set of features for data point.
The value is y sub i for each data point

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which is in this m by one vec, column
vector y.

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And, we want to make, try to find f such
that x times f is almost what?

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I'm going to use bold y and bold f to
indicate that they're vectors,

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A small mistake.
I've used bold y here it actually should

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be unbold y and that was a typo, I
apologize for that.

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Let's go on, though.
We want to find f at approximately matches

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this equation.
And we're going to do that by minimizing

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the difference between X transposed f and
yi for each of these data points by taking

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the sum of the squares.
That's what we decided earlier.

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And if we write the sum of squares in
matrix form, it turns out we simply, the

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difference between x, f and y transposed
and multiplied by itself.

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So, essentially, it's the norm or the sum
of squares of this difference vector,

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which is exactly what we have here.
This quantity needs to be minimized.

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Now, to minimize a quantity, if you go
back to high school, you need to make its

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derivative zero.
What derivative with respect to what?

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The unknown quantity is f,
In this case it's a vector.

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So, it's a bit more complicated than high
school.

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You want to take the derivative with
respect to every element of x.

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Well,
If you study the vector calculus, it's

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fairly easy to do that.
We won't go through that here. But, just

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think about f as being one variables for
the time being. the way you take the

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derivative, derivative is you would take
the, you would expand this out and you

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will get one term which is f transposed, X
transposed Xf,

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So you get two f's there. So, when you
take derivative of that, you will get

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twice X transpose Xf.
And then, you'll get two elements where

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there is a y transposed X into f, or an f
transposed X transposed into y.

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And, for each of those, you'll get a y
transposed there.

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So,
Of you set the derivative to zero, you'll

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get twice of this being zero which
essentially, this minus this being zero

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which are called the normal equations.
X transposed f times f should be equal to

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X transposed y.
Now, X transposed f, if you look at it in

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matrix form, it's actually no longer a
very large matrix.

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It's only an n by n matrix, because you
took this long matrix multiplied by its

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transposed you'll get an n by n matrix.
You have f which is a vector of length n,

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And X transpose y again will become a much
smaller column vector.

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And so, this, those two equations is n by
n, and this side is n, so you should be

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able to solve it exactly as long as, of
course, the matrix is not singular or it

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actually has a unique solution.
Most of the time, it does have a unique

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solution.
And so, we solve this to get our f.

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Once we have our f, our least squares
estimate for f is nothing but X transposed

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f,
Or x prime transposed f, where x prime is

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x, one.
So, given an x, the value of y most likely

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by least squares is, is x, one transpose
x, and that's our least squares estimate.

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Let's do an example.
Let's take a simple four data point

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example with only one feature.
So, x is just a one variable.

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X can be many variables, of course.
And, y is a set of values as well.

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We saw the normal equations,
Normal equations will look something like

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this.
If you take x transpose x, each element of

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x is x, 1..
Y is just the values of y.

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The f that you get by solving the normal
equation turns out to be this one.

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And when you plot f transposed x, which is
simply the line 0.11x - 0.26 on a graph,

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And you have the y,.
X, y coordinates being flooded as well,

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X, y.
You see that the line actually almost fits

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the, the, the data.
This is all that it, that there is in

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linear regression.
Of course, if there are many variables,

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you will have, well, possibly, many such
plots.

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One for each value, each, each, each
feature x,

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And, different, the f will be much longer
in terms of the number coefficients.

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So, we have a approximation for y for
every value of x that we might come

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across.
But, we also might want to ask, how good

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is this fit to the data.
A common measure of how good the fit is,

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is called the R squared value.
Which is simply the sum of squares of the

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errors of our actual estimates.
So, you have fx) of x, which is f

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transposed xy) sub i for each data point,
Minus the actual value. Square it and sum,

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And you'll divide that by the variation in
y about it's means.

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So, y has a certain mean which would be
somewhere here.

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And you want to see how the error, the sum
of all these errors compares with the

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actual variation in the y values across
their mean.

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Doesn't sound like a great measure
because, you know, if you, if you have,

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have a steeper slope, you, you probably
have a higher value over here. But then,

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if you have a steeper slope, you're
actually tolerating more error in your in

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your estimate.
But this is a common and easy-to-calculate

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measure.
There are better measures which tell you

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how good each of these coefficients are in
terms of how confident one should be in

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each coefficient.
But, we will not go into that in detail.

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For this particular example, the data is
pretty much fitting a line.

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The R square value comes out to be
somewhere around 0.95. As you can see, R

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squared value can be close to one, image
of this is a good fit.

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And can be close to zero if there isn't a
good fit.

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Let's take, take another example.
This is an example which is a little bit

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more realistic.
It comes from a book called Super

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Crunchers written by Ian Aryes in 2007.
It talks about the power of reasoning from

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data.
It's a great book if you want to read it.

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He, he tells the story of a wine expert
called Orley Ashenfelter who predicted the

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quality of wine based on the winter
rainfall, the average temperature, and the

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harverst rainfall in a season.
So, Ashenfelter could predict how good a

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particular wine would be depending on the
weather in the growing region for that

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wine and do this much before the wine hit
the market.

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And, turns out that his estimates were
simply based on linear regression and

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turned out to be extremely successful.
And, surprised many experienced wine

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critics who would have to taste the wine
many times and judge it before predicting

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whether it be high priced wine or not.
The kind of estimate he got was something

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like, you know, linear least squares fit
which is you know, the f0, the f1, the f2,

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and the f3.
And he got a very nice fit.

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The important thing to note is that, we
have possibly positive correlations

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between the output variable y and the
input variable, such as a positive value

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here, here, and here, or we can have a
negative correlation.

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So, in this case, 0.26 is negative,
0.00386 is negative.

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So, if harvest rain rainfall goes higher,
the quality goes down unlike the positive

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correlation of with respect to say, winter
rainfall.

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Let's take a look at a few more examples
of, of correlation that one might get with

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a single variable.
Of course, all this applies to multiple

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variables.
You can always draw graphs of the value

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with respect to any of the feature
variables.

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This is clearly very strong correlations.
You will get a very high R squared value

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for the data that looks like this.
Of course, if it looks a bit more

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scattered, you'll get a lower value via R.
Squared. And similarly, if it's negatively

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correlated, your slopes will be negative
which could have lower R squared if they

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are more scattered.
And, suppose your data is like, like this.

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Well, it doesn't look like there's any
correlation.

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00:10:11,522 --> 00:10:16,469
And the R2 squared will reflect because if
you look at the variation of y across its

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mean, it's fairly large.
And whatever line one puts around it, the

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variation of the y's with respect to
whatever line one decides, decides to draw

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will also be very large.
And so, this ratio will be close to one

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and R2 squared will be close to zero.
However,

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Let's look at this case for example.
This is a little bit more subtle. y is

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almost constant as x varies.
So, there's a line which goes through

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00:10:46,049 --> 00:10:48,983
these points.
But, what's the R squared value?

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00:10:48,983 --> 00:10:52,986
Y doesn't change at all,
And the line, if you fit it, might be

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exact or very close to being exact, so the
error might be very small as well.

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So, this ratio is again very close to one,
and the R squared value is very small.

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00:11:03,391 --> 00:11:06,860
So, essentially you're saying there's no
correlation.

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00:11:07,440 --> 00:11:12,769
And that's actually true because whatever
be the value of x, y doesn't change.

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00:11:12,769 --> 00:11:17,337
So, the only way you can get
non-correlation is not just if it, the

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00:11:17,337 --> 00:11:22,251
data is scattered all across,
But even if it is a straight line here. If

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00:11:22,251 --> 00:11:26,820
y doesn't change its vector x, then there
is no correlation either.

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And lastly, we have another situation
which we'll come to in the next segment

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00:11:33,700 --> 00:11:41,101
which is a situation like this.
Whatever line one draws through these

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00:11:41,101 --> 00:11:44,002
points, one will always make a lot of
error.

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This is an example of non-linear
correlation.

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00:11:47,438 --> 00:11:53,546
The data is correlated so you can draw a
line which is like a parabola it could

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00:11:53,546 --> 00:11:57,974
certainly work, but it doesn't work for
linear correlation.

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For this and many other reasons as we will
come to in the next segment, we have to go

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beyond linear least squares to more
complicated prediction models.
