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Nuts and Bolts cont.

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Nuts and Bolts cont.

1. Calculus Review

2. Random Variables and Probability Density Functions (PDF’s)

3. Cumulative Distribution Functions (CDF’s)

- Differentiation

- Partial Differentiation

- Basic Tools of Integration

- Multiple Integration

The derivative of a function f is the function f’ defined by:

- In class, we may also use the notation dy / dx.
- - Thinking carefully about the equation, we see that the derivative provides the measure of the slope of a function.
- To see this, slope = y/x
- Rewritten in terms of the above equation, slope = f(x) /x
- The change in f due to an arbitrarily small shift in x (as measured by h) is f( x + h ) - f( x )
- The change in x is measured by h.

Constant Rule: If f(x) = k, then f’(x) = 0.

e.g. Suppose f(x) = 3. What is f’(x)?

Power Function Rule. If f(x) = cxn, then f’(x) = cnxn-1

e.g. Suppose f(x) = 3x2, what is f’(x)?

Sum-Difference Rule.

If f(x) = g(x) ± h(x), then f’(x) = g’(x) ± h’(x)

e.g. Suppose f(x) = x2 + 3x3, what is f’(x)?

e.g. Suppose f(x) = 17 – 4x, what is f’(x)?

Product Rule:

If f(x) = g(x)h(x), then f’(x) = g’(x)h(x) + g(x)h’(x)

e.g. Suppose f(x) = (4x3)(5-x2). What is f’(x)?

Quotient Rule:

If f(x) = g(x) / h(x), then f’(x) = [g’(x)h(x)-g(x)h’(x) ] / h(x)2

e.g. Suppose f(x) = 2x2 / (x-2). What is f’(x)?

Log Rule: If f(x) = ln( g(x) ), then f’(x) = g’(x) / g(x)

e.g. Suppose f(x) = ln(x). What is f’(x)?

e.g. Suppose f(x) = log(3x + x2). What is f’(x)?

Exponential-Function Rule:

If f(x) = eg(x), then f’(x) = g’(x)eg(x)

e.g. Suppose f(x) = e3x, what is f’(x)?

Suppose you have a function y = f(x1,…,xn). Then the partial derivative is the equation:

The partial derivative of f with respect to x1 is the effect of an infinitely small shift in x1 (again measured by h) on f.

- It is the multivariate extension of the concept of the slope.

To compute a partial derivative with respect to x1, simply treat x2, …, xn as constants, and apply the standard rules of differentiation

Identify the partial derivatives of f with respect to x1 and x2 for the following equations?

Example 1. f(x1,x2) = 4x12 + 3x2.

Example 2. f(x1,x2) = ln(x1x2).

Whereas the concept of a derivative stemmed from the need to compute the slope of a function f(x), integral calculus emerged from the need to identify the area between a function f(x) and the x-axis.

For example, suppose you wanted to know the area under the function f(x) = 2 on the range from 0 to 10. Then, a numerical solution for this integral obviously exists, equals to 20.

f(x)

2

x

0

10

The integral of f(x) is defined as F(x) = f(x) dx.

For reasons that remain obscure to me, F(x) represents the “anti-derivative” of the function f(x). In other words,

F’(x) = f(x)

Therefore, a set of rules symmetric to those we used for differentiation apply to integration.

Rule 1) a dx = ax + c

e.g. What is 2 dx = 2x + c ?

Rule 2) a f(x) dx = a f(x) dx

e.g. What is 17 x3 dx ?

Rule 3) (u+v) dx = u dx + v dx = ux + vx + c

e.g. What is (4+13) dx ?

Rule 4) xn dx = xn+1 / (n+1) + c

e.g. What is x3 dx ?

Note that for each of these rules, we must add a constant of integration.

To compute the area under the curve between two boundaries, we substitute the maximum boundary into the equation and subtract that number from the quantity found when we substitute the minimum boundary into the equation.

Sometimes, rather than computing the area under a curve, you will want to compute the volume, or even the hyper-volume. In these cases, you will need to integrate over multiple dimensions.

For a function f(x,y), it’s integral F(x,y) is defined:

F(x,y) = f(x,y) dx dy

To solve for this double integral is pretty easy conceptually. It is legitimate to rewrite the equation:

F(x,y) = f(x,y) dx dy = [ f(x,y) dx] dy

So, all you have to do is compute the integral of f(x,y) with respect to x and then with respect to y. If there are limits of integration for x, these can be substitute in after the first integral is calculated to make the calculation even easier.

And for a function f(x, y, … , z), the integral F(x,y,…,z) is defined: F(x,y,…,z) = … f(x,y…,z) dx dy…dz

What is the multiple integral of f(x,y) = x2y ?

Answer: f(x,y) = x2y dx dy ?

x2y dx dy = [x3y/3 + c] dy = x3y2/6 + cy + d

- Definition of a random variable
- PDF’s
- CDF’s

Let S define the sample space for an experiment.

A random variable is a real-valued function that is defined on the space S.

This definition is obtuse. It simply means that each element of S can be represented by some number.

note that for a real-valued function y = f(x), real-valued just means that y is a number between - and .

Example: If your experiment was to flip a fair coin, the sample space would be heads or tails. A random variable X would be the assignment of 1 to the heads outcome and 0 to the tails outcome.

If we have a probability distribution over the sample space, we may also have a probability distribution for the random variable.

Example 1: If we flip a fair coin once, our sample space is a head (X=1) or a tail (X=0). The probability distribution states that Pr(X=1)=.5 and Pr(X=0)=.5

Example 2: If we flip a fair coin twice, there are four possible outcomes in our sample space. The probability distribution states that Pr(X=2)=.25; Pr(X=1)=.5; Pr(X=0)=.25.

For a discrete random variable X, the pdf of X is the function f such that for any x:

f(x) = Pr(X = x).

In other words, the pdf states the probability of observing each possible value of X. If x is not a possible value for X, then f(x) = 0.

Example: Suppose that the random variable X represents a fair coin toss where the probability of heads (X=1) = .5. Then the pdf of X is:

f(x) = .5x (.5)1-xfor x = 1,0

f(x) = 0otherwise

For a continuous random variable X, the pdf of X is the function f such that for any x:

f(x) = Pr(X = x).

If x is outside the sample space, Pr(x) = 0.

A peculiar property of pdf’s is that the probability of any x equals zero. Why?

Consequently, we must identify the probability of a range of values of X. Thus,

Pr(XA) = A f(x) dx

Further, we know that : Real Line f(x) dx = 1

Example 1. Suppose that X is a random variable that follows a uniform pdf between zero and one.

What is the pdf of X?

Example 2. Suppose that X is a random variable that follows a uniform pdf between zero and five.

What is the pdf of X?

Example 1. Suppose that X is a random variable that follows a uniform pdf between zero and one.

f(x) = 1if 0 x 1

f(x) = 0otherwise

Example 2. Suppose that X is a random variable that follows a uniform pdf between zero and five.

f(x) = 1/5if 0 x 5

f(x) = 0otherwise

The cumulative distribution function F of a random variable X is a function defined for each x as follows:

F(x) = Pr( X x )

The cdf simply states the probability that the random variable takes a value less than x.

Properties of the CDF (draw figures to illustrate)

P1. If x1 < x2, F(x1) < F(x2)

P2. For any given x, Pr(X > x) = 1 – F(x)

P3. For any given x, Pr(x1 < X < x2) = F(x2)- F(x1)

Example: Suppose that the random variable X represents a fair coin toss where the probability of heads (X=1) = .5.

What is the cdf of X?

Example: Suppose that the random variable X represents a fair coin toss where the probability of heads (X=1) = .5.

What is the cdf of X?

F(x) = 0if x < 0

F(x) = .5if 0 x < 1

F(x) = 1if x 1

1

X

0

-1

0

1

2

The CDF for a discrete random variable X can be written:

F(xj) = ij f(xi)

This is the pdf for X

Because of the use of summations, it is clear that the cdf for a discrete random variable is discontinuous.

The CDF for a continuous random variable X can be written:

F(x) = - to x f(t) dt

This is the pdf for X

Notice how the notation has become cumbersome, with t’s standing in for x’s.

Example. Suppose that X is a random variable that follows a uniform pdf between zero and five.

f(x) = 1/5if 0 x 5

f(x) = 0otherwise

What is the cumulative distribution function F(x)?

What is the probability that X < 2 ?

Example. Suppose that X is a random variable that follows a uniform pdf between zero and five.

f(x) = 1/5if 0 x 5

f(x) = 0otherwise

What is the cumulative distribution function?

F(x) = 0 to X 1/5 dt

F(x) = t/5 |0 to X = x/5 – 0 = x/5

What is the probability that X < 2? F(2) = 2/5 = .4