Question

A random variable X has the probability density function given below

g(x)= {2x,0≤x<1
{ 6-x,1≤x<2
{ 0, otherwise

Show graphically and find out the probabilities: P(X≤3)

Answer 1

$g(x) = \begin{cases} 2x, & 0 \leq x<1 \\ 6-x, & 1\leq x <2 \\ 0, & otherwise\end{cases}$

Plotting this, we get following

However, in order to be indeed probability density function condition $\int_{-\infty}^{+\infty} g(x)dx =1$ must be fulfilled. In our case,

$\int_0^1(2x)dx + \int_1^2(6-x)dx = x^2|_0^1 + 6x|_1^2 - \frac{x^2}{2}|_1^2 = 1 +6 -2 - 0.5 = 4.5.$

In such case, g(x) should be normalized, namely $\tilde{g}(x) = g(x)/4.5$.

In order to find $P(X \leq3)$ we need to integrate $\tilde{g}(x)$ from $-\infty$ to 3 or, what is the same, find the area below the (normalized) plot. Since for $x \geq 3$ $\tilde{g}(x) =0$, integration from $-\infty$ to 3 is the same as integration from $-\infty$ to $+\infty$. The last one we already did above.

$P(X \leq 3) = \int_{-\infty}^3 g(x)dx =\frac{\int_{-\infty}^3 \tilde{g}(x)dx }{4.5} = \frac{\int_{-\infty}^{+\infty}\tilde{g}(x)dx }{4.5} = 1$.

Alternative approach: $P(X \leq 3) = 1 - P(X \geq 3) = 1$, since for $x \geq 3$ $g(x) =0$ and integral of it is 0 too.

Answer: $P(X \leq 3) = 1.$