Question

for a random variable X that is triangularly distributed over the interval (a,b). That is, f_X(a) = 0, f_X(b) = 0, and the peak value for f_X occurs at (a+b)/2.show the plots of the probability density function and the cumulative distribution function?

Answer 1

The random variable X follows a triangular distribution over the interval (a, b). Let f_X be the pdf of X. Hence, given the bounds, we have:

The peak value of the pdf occurs at (a + b) / 2. Therefore, the pdf(X) increases linearly from a to (a + b) / 2 and then decreases linearly from (a + b) / 2 to b.

Plotting the pdf, we will get a triangle like this:

The center of the pdf lies at (a + b) / 2, which is midway between a and b. Hence, the triangle is an isosceles triangle with the median from the top point on the bottom line also serving the role of the height of the triangle. The area of the triangle is given using the formula:

Here, height is the f(X) at (a + b) / 2 and Base is the distance (b - a). Since, f(X) is a pdf, the area of the triangle will be 1. Hence, we have:

Thus, our pdf becomes:

The cdf is the cumulative frequency distribution and is obtained by summing all the previous values of the pdf.

Hence, cdf(X) can be represented as F(X) and will be given as:

Therefore, the integral of the pdf becomes the cdf. Initially, the pdf rises linearly and then decreases linearly and hence the sum or the integral of pdf will follow a sigmoid as shown below: