g ( x ) = { 2 x , 0 ≤ x < 1 6 − x , 1 ≤ x < 2 0 , o t h e r w i s e g(x) = \begin{cases} 2x, & 0 \leq x<1 \\ 6-x, & 1\leq x <2 \\ 0, & otherwise\end{cases} g ( x ) = ⎩ ⎨ ⎧ 2 x , 6 − x , 0 , 0 ≤ x < 1 1 ≤ x < 2 o t h er w i se
Plotting this, we get following
However, in order to be indeed probability density function condition ∫ − ∞ + ∞ g ( x ) d x = 1 \int_{-\infty}^{+\infty} g(x)dx =1 ∫ − ∞ + ∞ g ( x ) d x = 1 must be fulfilled. In our case,
∫ 0 1 ( 2 x ) d x + ∫ 1 2 ( 6 − x ) d x = x 2 ∣ 0 1 + 6 x ∣ 1 2 − x 2 2 ∣ 1 2 = 1 + 6 − 2 − 0.5 = 4.5. \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. ∫ 0 1 ( 2 x ) d x + ∫ 1 2 ( 6 − x ) d x = x 2 ∣ 0 1 + 6 x ∣ 1 2 − 2 x 2 ∣ 1 2 = 1 + 6 − 2 − 0.5 = 4.5.
In such case, g(x) should be normalized, namely g ~ ( x ) = g ( x ) / 4.5 \tilde{g}(x) = g(x)/4.5 g ~ ( x ) = g ( x ) /4.5 .
In order to find P ( X ≤ 3 ) P(X \leq3) P ( X ≤ 3 ) we need to integrate g ~ ( x ) \tilde{g}(x) g ~ ( x ) from − ∞ -\infty − ∞ to 3 or, what is the same, find the area below the (normalized) plot. Since for x ≥ 3 x \geq 3 x ≥ 3 g ~ ( x ) = 0 \tilde{g}(x) =0 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 ≤ 3 ) = ∫ − ∞ 3 g ( x ) d x = ∫ − ∞ 3 g ~ ( x ) d x 4.5 = ∫ − ∞ + ∞ g ~ ( x ) d x 4.5 = 1 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 P ( X ≤ 3 ) = ∫ − ∞ 3 g ( x ) d x = 4.5 ∫ − ∞ 3 g ~ ( x ) d x = 4.5 ∫ − ∞ + ∞ g ~ ( x ) d x = 1 .
Alternative approach: P ( X ≤ 3 ) = 1 − P ( X ≥ 3 ) = 1 P(X \leq 3) = 1 - P(X \geq 3) = 1 P ( X ≤ 3 ) = 1 − P ( X ≥ 3 ) = 1 , since for x ≥ 3 x \geq 3 x ≥ 3 g ( x ) = 0 g(x) =0 g ( x ) = 0 and integral of it is 0 too.
Answer: P ( X ≤ 3 ) = 1. P(X \leq 3) = 1. P ( X ≤ 3 ) = 1.
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