Question

a) suppose that a spinner returns a real number between zero and one. What would the height of the PDF function be for this random variable X. Sketch its graph and make sure the total area under the graph is 1.

b) Now suppose that the random variable of interest is a number between 1 and 3. We will obtain this number by creating a new random variable Y=2X + 1 . Explain how you know Y will give a number in the correct range. What would be the height of Y’s PDF. Sketch it’s graph and make sure the total area under the graph is 1.

c) finally consider any random variable of the form W = a +cX, where a & c are real numbers and c does not equal zero. what range of values will this variable assume? What would be the height of its PDF? Sketch it’s graph, and make sure the total area under the graph is 1.

Answer 1

a) Since the spinner returns a real number between zero and one, we assume uniformity and then the random variable X  is Uniform(0,1). Then the height of the pdf will be unity over the entire range [0,1].

From the graph, we find that the total area of the region described by the red square of sides unity=height *length=1

b) Note that x<=1 implies and implied by , 2x+1<=2*1+1=3

Also x>=0 implies and implied by 2x+1>=0+1=1. Combining we get 0<=x<=1 implies and implied by 1<=y<=3. Hence y gives values in the correct range (1,3).

Since, Y~Uniform(1,3), the density of Y is f(y)=1*1/2 .  Then the height of the pdf of Y is 0.5.

From the graph, we find that the total area of the region described by the red rectangle with length =1 and height=.5 is 2*.5=1.

c) Suppose c>0, then a+cX>=a and  then a+cX<=a+c so that the range of W is [a,a+c].

For c<0, then a+cX<=a and  then a+cX>=a+c so that the range of W is [a+c,a].

Then the height of the PDF will be 1/|c|, considering both c>0 and c<0.

Assuming c>0, from the graph, we find that the total area of the region described by the red rectangle with height =1/c and length=(c+a-a)=c is c/c=1.