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Uniform Probability Model

The probability model of a discrete random variable was evident to you before you took this course. Sure, the notation is new but in playing games like craps you knew the probabilities of interest; eg, probability of a "7". Recall that in general, the probability model of a discrete random variable with range $\{1,2, ..., k\}$ consists of probabilities P(X = i) for i = 1,2, ..., k. The probability of a continuous random variable is not as evident. Lets begin with an example where we know the answers. This will motivate the continuous probability model.

Suppose we choose a real number at random between 0 and 1. Let X be the number chosen. Then X is a continuous random variable with the interval (0,1) as its range. Certain probabilities are obvious here:

1.
The probability that X is between 0 and 1/2 is 1/2.

2.
The probability that X is between 1/2 and 1 is 1/2.

3.
The probability that X is between 0 and 1/4 is 1/4.

4.
The probability that X is between 1/2 and 3/4 is 1/4.

5.
The probability that X is between 1/8 and 2/8 is 1/8.
Are you ready for the big jump? What's the probability that X is between a and b when a and b are real numbers between 0 and 1? It's b - a, the length of the interval. Go back to the list above and check if this isn't so for those cases. This leads to the following, though:
1.
The probability that X is between 1/4 and 3/4 is 1/2.

2.
The probability that X is between 3/8 and 5/8 is 1/4.

3.
The probability that X is between 7/16 and 9/16 is 1/8.

4.
The probability that X is between 15/32 and 17/32 is 1/16.
Note that we could continue this list forever. Each of the above intervals contains the number 1/2 and that further the length of each succeeding interval is getting smaller. Hence, we must have P(X = 1/2) = 0. But this is true for any real number a between 0 and 1; i.e., P(X = a) = 0. In general, for continuous random variables the discrete probability model will not work. But the probabilities of intervals are the probabilities of interest and this is how we define the model.

For a continuous random variable X whose range is the interval (c,d) the probability model  of X is a curve f(x) such that the probability that X is between a and b is the area under the curve between a and b, that is, for some function f(x) the P(a < X < b) is the area under the curve as shown in Figure 5.1


  
Figure 5.1: Area under the curve
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\begin{center}
\epsfig{file=fig16.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Notice that f(x) cannot be negative and that the total area under the curve must be one.

Consider the above example where X is a number chosen at random between 0 and 1. If we draw a straight line with slope 0 and height 1 above the interval (0,1), then the area under this line over the interval (a,b) is $(b - a) \times 1 = b-a$, which is our desired probability. The curve is given in Figure 5.2. This is called the uniform probability model .


  
Figure 5.2: Uniform(0,1)
\begin{figure}
\begin{center}
\epsfig{file=fig17.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Here's a second example. Suppose we choose a point at random inside the unit circle, a circle with radius 1 and center at the origin, (sketch it!).


  
Figure 5.3: Circle with radius 1
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\begin{center}
\epsfig{file=fig18.ps, height=5in, width=6in, angle= -90}\end{center}\end{figure}

Let X be the distance between the chosen point and the origin, (sketch it!). See my sketch in Figure 5.3. Then the range of X is between 0 and 1, just like the uniform. But the probabilities are unlike the uniform. For example, it is much more likely that X is between 3/4 and 1 than between 0 and 1/4, (Why? Sketch it!). In fact, you can show that the probability model for X is a line over the interval (0,1) with intercept at the origin and slope 2 (a triangle!, sketch it!). The graph is


  
Figure 5.4: Triangular distribution
\begin{figure}
\begin{center}
\epsfig{file=fig19.ps, height=5in, width=5in, angle= -90}\end{center}\end{figure}

Recall that the area of a triangle is $\frac{1}{2} \times$ base $\times$ height. Show that the area of the triangle is 1. Next shade in the area under the curve from 1/4 to 3/4. Determine this area. You have just found P(1/4 < X < 3/4).


Exercise 6.1.1  
1.
Let X be a number chosen at random between 8 and 10. Determine the probabilities that X is between 8.5 and 9.5 and X is between 8.5 and 8.7. Determine the probability that X is between a and b. Verify that the probability model for X has the graph given in Figure 5.5.


  
Figure 5.5: Uniform(8,10)
\begin{figure}
\begin{center}
\epsfig{file=figex51.ps, height=5in, width=5in, angle=-90}\end{center}\end{figure}

2.
For the second example above, find the probability that X is between 0 and 1/2.
3.
For the second example above, find the probability that X is between 3/4 and 1.
4.
For the second example above, find the probability that X is between 0 and 1/4.
5.
For the second example above, (choose a point at random inside the unit circle), find c so that the probability that X is between 0 and c is 1/2. (Hint Sketch it!).


next up previous contents index
Next: Parameters Up: Continuous Probability Models Previous: Continuous Probability Models

2001-01-01