Science:MATH105 Probability/Lesson 2 CRV/2.06 A Simple PDF Example

Question

Let f(x) = k(3x² + 1).

1. Find the value of k that makes the given function a PDF on the interval 0 ≤ x ≤ 2.
2. Let X be a continuous random variable whose PDF is f(x). Compute the probability that X is between 1 and 2.
3. Find the distribution function of X.
4. Find the probability that X is exactly equal to 1.

Solution

Part 1)

${\begin{aligned}1&=\int _{0}^{2}f(x)dx\\&=\int _{0}^{2}k(3x^{2}+1)dx\\&=k({\frac {3x^{3}}{3}}+x){\Big |}_{0}^{2}dx\\&=k(10)\end{aligned}}$

Therefore, k = 1/10.

Notice that f(x) ≥ 0 for all x. Also notice that we can rewrite this PDF in the obvious way so that it is defined for all real numbers:

$f(x)={\begin{cases}{\frac {1}{10}}(3x^{2}+1),&{\text{if }}0\leq x\leq 2\\0,&{\text{otherwise}}\end{cases}}$

Part 2)

Using our value for k from Part 1:

${\begin{aligned}\mathrm {Pr} (1\leq X\leq 2)=\int _{1}^{2}{\frac {3x^{2}+1}{10}}dx={\frac {x^{3}+x}{10}}{\Big |}_{1}^{2}=1-2/10=4/5\end{aligned}}$

Therefore, Pr(1 ≤ X ≤ 2) is 4/5.

Part 3)

Using the Fundamental Theorem of Calculus, the CDF of X at x in [0,2] is

${\begin{aligned}{\text{Pr}}(X\leq x)=F(x)&=\int _{-\infty }^{x}f(t)dt\\&=\int _{0}^{x}{\frac {1}{10}}(3t^{2}+1)dt\\&={\frac {1}{10}}(t^{3}+t)|_{0}^{x}\\&={\frac {1}{10}}(x^{3}+x),{\text{ for }}0\leq x\leq 2\end{aligned}}$

We can also easily verify that F(x) = 0 for all x < 0 and that F(x) = 1 for all x > 2.

Part 4)

Since X is a continuous random variable, we immediately know that the probability that it equals any one particular value must be zero. More directly, we compute

${\text{Pr}}(X=1)=\int _{1}^{1}f(t)dt=0$