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CaltechX - SP | bkFXRWrmn9M
RANGEL: Now consider a graphical intuition for why the second order conditions look like they
do. To remind you, the second order condition is given by the second derivative,
and the local optimum has to be less than 0. That provides, together with the first
order conditions, sufficient conditions for optimization.
Now, to see why that has to be the case, compare the following two functions. The
one on the left looks like this, and it doesn't satisfy the second order conditions of
the point x star. In fact, it is easy to see that, in this case, the second derivative at x
star is greater than 0. The slope is becoming progressively more positive as you
move to the right.
Compare this to a function that looks like this-- just like the one before-- and at
which at x star, our candidate for the local optimum, the second derivative, satisfies
the second order condition. Let's see why it has to be the case that x star is not a
maximum here, but is a maximum here, and the key nature of the difference is
captured by the second order condition.
Now, if you start at x star in the field on the left and move a tiny bit to the right by
delta x, since the slope is 0, you don't really change the value of the function. So
that doesn't help you. However-- and this is the key intuition that you have to see--
because the second derivative is positive, the value of the slope changes when you
move from x star to x star plus delta x and becomes positive here at the new point.
That means that if you carry out a farther move-- let's say by another delta x-- since
the slope was positive at x star plus delta x, the function now has increased. And
that means that by moving 2 delta x to the right, you can improve the value of thefunction, which means that it cannot be an optimum.
In contrast, see what happens if you try to play the same game at x star in the field
on the right. As before, the first order conditions are satisfied, so the slope is 0 at x
star. But if you try to move to the right because -- let's say by delta x-- because the
second derivative is now less than 0, the slope actually goes down and it becomes
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negative. Therefore, if you try to move farther the x, the value of the function goes
down.
It should be easy to see, and you should convince yourself, that it doesn't matter,
for these arguments, whether the movement is to the right or to the left. In both
cases, they go through in analogous way. This demonstrates the intuition for why
the second order condition T double prime of x star less than 0 provides sufficient
conditions for the point to be a local maximum.
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