Module 6: Probability and Simulation

• lim_n→infty y_n
• lim_n→infty z_n
• lim_n→infty F(y_n)
• lim_n→infty F(z_n)

• Argue that Pr[X>k] = (1-p)^k.
• Then write down the CDF.
• Write down an expression for Pr[3.11 < X ≤ 5.43] in terms of the CDF.

Exercise 22: Recall the Bus-stop example. Suppose rv X denotes the first interarrival time. Download BusStop.java and BusStopExample.java. Then modify the latter to estimate the following:

• Pr[X < 1]
• Pr[X < 1.1]
• Pr[X=1]

Exercise 23: What is an example of a continuous rv associated with the QueueControl.java application?

The difficulty with continuous rv's:

• First, Pr[X=x] = 0 for most continuous rv's X.
       => This means we can't specify the distribution using a list.

• Second, it's not clear how we can get probabilities to sum to 1.
       e.g., for a rv that takes any real value in [0,1], how do we sum probabilities?

• Fortunately, the CDF offers a way out.

Let's write some code to estimate a CDF:

• First, identify the range [a,b] of interest.

• Then, break this up into M small intervals:
       [a,a+δ], [a+δ,a+2δ], ..., [a+(M-1)δ,a+Mδ]

• Maintain a counter intervalCount[i] for each interval.

• For any random value x, identify the interval
       j = floor ( (x-a) / δ )
  and increment the count of every interval j ≥ i.

• Pseudocode:

      Algorithm: cdfEstimation (M, a, b)
      Input: number of intervals M, range [a,b]
      1.   δ = (b-a) / M
      2.   for n=1 to numTrials 
      3.       x = generateRV ()
      4.       j = floor ( (x-a) / δ )
      5.       for i=j to M-1
      6.           intervalCount[i] = intervalCount[i] + 1
      7.       endfor
      8.   endfor
           // Now build and print the CDF
      9.   for i=0 to M-1
      10.      midPoint = a + iδ + δ/2                           // Interval: [a+iδ,a+(i+1)δ]
      11.      print (midPoint, intervalCount[i] / numTrials)
      12.  endfor
      

• Sample code: (source file)

      static Function makeUniformCDF ()
      {
  	double a = 0, b = 1;
  	int M = 50;                   // Number of intervals.
  	double delta = (b-a) / M;     // Interval size.
  
  	double[] intervalCounts = new double [M];
  	double numTrials = 1000000;
  
  	for (int t=0; t<numTrials; t++) {
              // Random sample:
  	    double y = RandTool.uniform ();
              // Find the right interval:
              int k = (int) Math.floor ((y-a) / delta);
              // Increment the count for every interval above and including k.
              for (int i=k; i<M; i++) {
                  intervalCounts[i] ++;
              }
  	}
  
  	// Now compute probabilities for each interval.
  	double[] cdf = new double [M];
  	for (int k=0; k<M; k++) {
  	    cdf[k] = intervalCounts[k] / numTrials;
  	}
  
          // Build the CDF. Use mid-point of each interval.
          Function F = new Function ("Uniform cdf");
  	for (int k=0; k<M; k++) {
  	    double midPoint = a + k * delta + delta/2;
              F.add (midPoint, cdf[k]);
  	}
  
          return F;
      }
       

Getting probabilities out of a CDF:

• Recall that for any rv X with CDF F:
       Pr[y₁ < X ≤ y₂] = F(y₂) - F(y₁)

• Thus, for example, if X has the uniform CDF
       Pr[0.75 < X ≤ 1.0] = F(1) - F(0.75)

• Important: the CDF gets you the probability that X's outcome lands in any particular interval of interest:
       Pr[X falls in any interval (a,b]] = F(b) - F(a)

A small technicality that goes to the heart of a continuous rv:

• If you were observant, you would have noticed something peculiar about the statement
       Pr[X in any interval (a,b]] = F(b) - F(a)

• The interval is NOT written as [a,b] or (a,b) or [a,b].

• Instead, it's written as: (a,b]

• What does this mean?

  

  The interval (a,b] includes b but does NOT include a.

• Why is this relevant?

  

  • Because a discrete rv's CDF has jumps
         => We need to account for the jump points carefully.
  • We say that the CDF for a discrete rv is discontinuous.
  • A continous rv's CDF, on the other hand, is continuous (smooth, no jumps).
         => Inclusion or exclusion of an end point makes no difference.

• Another related point: consider lim_a→b F(b) - F(a)

  

  • For a continuous rv
         lim_a→b F(b) - F(a) = 0
  • This is why Pr[X=a] = 0
         Pr[X=a] = lim_b→a F(b)-F(a) = 0

• A few more observations:
  • A discrete rv's CDF is continuous in places
         In fact, it's discontinous only at the points where the jumps occurs
  • So, is this useful?

    

    
         No, because the non-jump points never actually occur
         => We are never interested in probabilities of intervals between jump points.
         => Those probabilities are always zero.

• Finally, since F is continuous for a continuous distribution, we might ask: is the derivative F' of any value?
       We'll explore this next.

About CDFs and derivatives:

• The concept of CDF applies to both discrete and continuous rv's.

• The CDF of a discrete rv is not differentiable.

• For continuous rv's, the derivative is called a probability density function (pdf).

• The pdf has two uses:
  • It's useful in some calculations as we'll see below.
  • Most books use the pdf
         => You need to be comfortable with the pdf to read them.

• For completeness, note that the list of probabilities for a discrete rv is called a probability mass function (pmf), e.g.
       Pr[X=0] = 0.064
       Pr[X=1] = 0.288
       Pr[X=2] = 0.432
       Pr[X=3] = 0.216

• Customarily, a discrete rv's distribution is specified by giving its pmf.
       => The CDF is not really useful for discrete rv's

• Customarily, a continuous rv's distribution is specified by giving its pdf.
       => The CDF is equivalent (because you can take its derivative to get the pdf)

• To summarize:
  • Discrete rv's use a pmf.
  • Continuous rv's use a pdf.
  • Both use a CDF.

Expectation - the discrete case