Question

Find the cdf \(F(x)\) associated with each of the following probability density functions. Sketch the graphs of \(f(x)\) and \(F(x)\). (a) \(f(x)=3(1-x)^{2}, 0

Short answer

The CDFs are \(F(x) = 1 - (1-x)^3\) for (a), \(F(x) = 1 - 1 / x\) for (b), and \(F(x) = x / 3\) for \(0 < x < 1\) and \(F(x) = (x - 2) / 3 + 1 /3\) for \(2 < x < 4\) in (c). The median and the 25th percentile for each distribution need to be calculated by solving the respective CDF equal to 0.5 and 0.25.

Step-by-step solution

Part a

Given the PDF \(f(x) = 3(1-x)^{2}\) for \(0 < x < 1\). To find the CDF, integrate from the lower limit to \(x\): \(F(x) = \int_{0}^{x} f(t) dt = \int_{0}^{x} 3(1-t)^{2} dt = [-(1-t)^3]|_{0}^{x} = 1 - (1-x)^3\). The graph of \(f(x)\) is a downward parabola while \(F(x)\) starts from 0 at \(x = 0\) and ends at 1 at \(x = 1\). Solve \(1 - (1-x)^3 = 0.5\) for the median and \(1 - (1-x)^3 = 0.25\) for the 25th percentile.

Part b

Given the PDF \(f(x) = 1 / x^{2}\) for \(1 < x < \infty\). The CDF is \(F(x) = \int_{1}^{x} f(t) dt =\int_{1}^{x} 1 / t^{2} dt = [-1/t] |_{1}^{x} = 1 - 1 / x\). The graph of \(f(x)\) is a downward hyperbola and \(F(x)\) starts at 0 at \(x = 1\) and ends at 1 as \(x -> \infty\). Solve \(1 - 1 / x = 0.5\) for the median and \(1 - 1 / x = 0.25\) for the 25th percentile.

Part c

Given the PDF \(f(x) = 1 / 3\) for \(0 < x < 1\) and \(2 < x < 4\). The CDF is \(F(x) = \int_{0}^{x} f(t) dt\) for \(0 < x < 1\) and \(F(x) = \int_{2}^{x} f(t) dt + 1 / 3\) for \(2 < x < 4\). The graphs of \(f(x)\) are straight lines at height 1/3 while \(F(x)\) raises linearly to 1/3 for \(x < 1\) and raises linearly to 1 for \(2 < x < 4\). Solve \(x/3 = 0.5\) for the median and \(x/3 = 0.25\) for the 25th percentile. Since the solutions are outside the range of \(x\), take \(\min(\max(x, 0), 1)\) where \(x\) is the solution.