Find the distribution function F ( x ) = ∫ − ∞ x f ( x ) d x F(x) = \int\limits_{ - \infty }^x {f(x)dx} F ( x ) = − ∞ ∫ x f ( x ) d x
If x < 2 x < 2 x < 2 F ( x ) = ∫ − ∞ x 0 d x = 0 F(x) = \int\limits_{ - \infty }^x {0dx} = 0 F ( x ) = − ∞ ∫ x 0 d x = 0
If x ≤ 3 x \le 3 x ≤ 3 F ( x ) = ∫ − ∞ 2 0 d x + ∫ 2 x 1 2 ( x − 2 ) d x = x 2 4 ∣ 2 x − x ∣ 2 x = x 2 − 4 4 − x + 2 = x 2 4 − x + 1 F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^x {\frac{1}{2}(x - 2)dx} = \left. {\frac{{{x^2}}}{4}} \right|_2^x - \left. x \right|_2^x = \frac{{{x^2} - 4}}{4} - x + 2 = \frac{{{x^2}}}{4} - x + 1 F ( x ) = − ∞ ∫ 2 0 d x + 2 ∫ x 2 1 ( x − 2 ) d x = 4 x 2 ∣ ∣ 2 x − x ∣ 2 x = 4 x 2 − 4 − x + 2 = 4 x 2 − x + 1
If x ≤ 5 x \le 5 x ≤ 5 F ( x ) = ∫ − ∞ 2 0 d x + ∫ 2 3 1 2 ( x − 2 ) d x + ∫ 3 x a d x = x 2 4 ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 x = 9 − 4 4 − 3 + 2 + a ( x − 3 ) = 5 4 − 1 + a x − 3 a = a x − 3 a + 1 4 F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^x {adx} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^x = \frac{{9 - 4}}{4} - 3 + 2 + a(x - 3) = \frac{5}{4} - 1 + ax - 3a = ax - 3a + \frac{1}{4} F ( x ) = − ∞ ∫ 2 0 d x + 2 ∫ 3 2 1 ( x − 2 ) d x + 3 ∫ x a d x = 4 x 2 ∣ ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 x = 4 9 − 4 − 3 + 2 + a ( x − 3 ) = 4 5 − 1 + a x − 3 a = a x − 3 a + 4 1
If x ≤ 6 x \le 6 x ≤ 6
F ( x ) = ∫ − ∞ 2 0 d x + ∫ 2 3 1 2 ( x − 2 ) d x + ∫ 3 5 a d x + ∫ 5 x ( 2 − b x ) = x 2 4 ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 5 + 2 x ∣ 5 x − b x 2 2 ∣ 5 x = 9 − 4 4 − 3 + 2 + 2 a + 2 ( x − 5 ) − b 2 ( x 2 − 25 ) = 5 4 − 1 + 2 a + 2 x − 10 − b x 2 2 + 25 b 2 = − b x 2 2 + 2 x + 2 a + 25 b 2 − 39 4 F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^5 {adx} + \int\limits_5^x {\left( {2 - bx} \right)} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^5 + 2\left. x \right|_5^x - \left. {\frac{{b{x^2}}}{2}} \right|_5^x = \frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(x - 5) - \frac{b}{2}({x^2} - 25) = \frac{5}{4} - 1 + 2a + 2x - 10 - \frac{{b{x^2}}}{2} + \frac{{25b}}{2} = - \frac{{b{x^2}}}{2} + 2x + 2a + \frac{{25b}}{2} - \frac{{39}}{4} F ( x ) = − ∞ ∫ 2 0 d x + 2 ∫ 3 2 1 ( x − 2 ) d x + 3 ∫ 5 a d x + 5 ∫ x ( 2 − b x ) = 4 x 2 ∣ ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 5 + 2 x ∣ 5 x − 2 b x 2 ∣ ∣ 5 x = 4 9 − 4 − 3 + 2 + 2 a + 2 ( x − 5 ) − 2 b ( x 2 − 25 ) = 4 5 − 1 + 2 a + 2 x − 10 − 2 b x 2 + 2 25 b = − 2 b x 2 + 2 x + 2 a + 2 25 b − 4 39
If x > 6 x > 6 x > 6 F ( x ) = ∫ − ∞ 2 0 d x + ∫ 2 3 1 2 ( x − 2 ) d x + ∫ 3 5 a d x + ∫ 5 6 ( 2 − b x ) = x 2 4 ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 5 + 2 x ∣ 5 6 − b x 2 2 ∣ 5 6 = 9 − 4 4 − 3 + 2 + 2 a + 2 ( 6 − 5 ) − b 2 ( 36 − 25 ) = 5 4 − 1 + 2 a + 2 − 11 b 2 = 9 4 + 2 a − 11 b 2 F(x) = \int\limits_{ - \infty }^2 {0dx} + \int\limits_2^3 {\frac{1}{2}(x - 2)dx} + \int\limits_3^5 {adx} + \int\limits_5^6 {\left( {2 - bx} \right)} = \left. {\frac{{{x^2}}}{4}} \right|_2^3 - \left. x \right|_2^3 + a\left. x \right|_3^5 + 2\left. x \right|_5^6 - \left. {\frac{{b{x^2}}}{2}} \right|_5^6 = \frac{{9 - 4}}{4} - 3 + 2 + 2a + 2(6 - 5) - \frac{b}{2}(36 - 25) = \frac{5}{4} - 1 + 2a + 2 - \frac{{11b}}{2} = \frac{9}{4} + 2a - \frac{{11b}}{2} F ( x ) = − ∞ ∫ 2 0 d x + 2 ∫ 3 2 1 ( x − 2 ) d x + 3 ∫ 5 a d x + 5 ∫ 6 ( 2 − b x ) = 4 x 2 ∣ ∣ 2 3 − x ∣ 2 3 + a x ∣ 3 5 + 2 x ∣ 5 6 − 2 b x 2 ∣ ∣ 5 6 = 4 9 − 4 − 3 + 2 + 2 a + 2 ( 6 − 5 ) − 2 b ( 36 − 25 ) = 4 5 − 1 + 2 a + 2 − 2 11 b = 4 9 + 2 a − 2 11 b
Then F ( x ) = { 0 , x < 2 x 2 4 − x + 1 , 2 ≤ x ≤ 3 a x − 3 a + 1 4 , 3 < x ≤ 5 − b x 2 2 + 2 x + 2 a + 25 b 2 − 39 4 , 5 < x ≤ 6 9 4 + 2 a − 11 b 2 , x > 6 F(x) = \left\{ {\begin{matrix}
{0,\,\,x < 2}\\
{\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\
{ax - 3a + \frac{1}{4},\,\,3 < x \le 5}\\
{ - \frac{{b{x^2}}}{2} + 2x + 2a + \frac{{25b}}{2} - \frac{{39}}{4},\,\,5 < x \le 6}\\
{\frac{9}{4} + 2a - \frac{{11b}}{2},\,\,x > 6}
\end{matrix}} \right. F ( x ) = ⎩ ⎨ ⎧ 0 , x < 2 4 x 2 − x + 1 , 2 ≤ x ≤ 3 a x − 3 a + 4 1 , 3 < x ≤ 5 − 2 b x 2 + 2 x + 2 a + 2 25 b − 4 39 , 5 < x ≤ 6 4 9 + 2 a − 2 11 b , x > 6
By the properties of the distribution function
lim x → 3 − 0 F ( x ) = lim x → 3 + 0 F ( x ) ⇒ 3 2 4 − 3 + 1 = 3 a − 3 a + 1 4 \mathop {\lim }\limits_{x \to 3 - 0} F(x) = \mathop {\lim }\limits_{x \to 3 + 0} F(x) \Rightarrow \frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \frac{1}{4} x → 3 − 0 lim F ( x ) = x → 3 + 0 lim F ( x ) ⇒ 4 3 2 − 3 + 1 = 3 a − 3 a + 4 1
lim x → 5 − 0 F ( x ) = lim x → 5 + 0 F ( x ) ⇒ 5 a − 3 a + 1 4 = − 25 b 2 + 10 + 2 a + 25 b 2 − 39 4 \mathop {\lim }\limits_{x \to 5 - 0} F(x) = \mathop {\lim }\limits_{x \to 5 + 0} F(x) \Rightarrow 5a - 3a + \frac{1}{4} = - \frac{{25b}}{2} + 10 + 2a + \frac{{25b}}{2} - \frac{{39}}{4} x → 5 − 0 lim F ( x ) = x → 5 + 0 lim F ( x ) ⇒ 5 a − 3 a + 4 1 = − 2 25 b + 10 + 2 a + 2 25 b − 4 39
lim x → 6 − 0 F ( x ) = lim x → 6 + 0 F ( x ) ⇒ − 36 b 2 + 12 + 2 a + 25 b 2 − 39 4 = 9 4 + 2 a − 11 b 2 \mathop {\lim }\limits_{x \to 6 - 0} F(x) = \mathop {\lim }\limits_{x \to 6 + 0} F(x) \Rightarrow - \frac{{36b}}{2} + 12 + 2a + \frac{{25b}}{2} - \frac{{39}}{4} = \frac{9}{4} + 2a - \frac{{11b}}{2} x → 6 − 0 lim F ( x ) = x → 6 + 0 lim F ( x ) ⇒ − 2 36 b + 12 + 2 a + 2 25 b − 4 39 = 4 9 + 2 a − 2 11 b
F ( ∞ ) = 1 ⇒ 9 4 + 2 a − 11 b 2 = 1 F(\infty ) = 1 \Rightarrow \frac{9}{4} + 2a - \frac{{11b}}{2} = 1 F ( ∞ ) = 1 ⇒ 4 9 + 2 a − 2 11 b = 1
We have the system of equations:
{ 3 2 4 − 3 + 1 = 3 a − 3 a + 1 4 5 a − 3 a + 1 4 = − 25 b 2 + 10 + 2 a + 25 b 2 − 39 4 − 36 b 2 + 12 + 2 a + 25 b 2 − 39 4 = 9 4 + 2 a − 11 b 2 9 4 + 2 a − 11 b 2 = 1 \left\{ \begin{array}{l}
\frac{{{3^2}}}{4} - 3 + 1 = 3a - 3a + \frac{1}{4}\\
5a - 3a + \frac{1}{4} = - \frac{{25b}}{2} + 10 + 2a + \frac{{25b}}{2} - \frac{{39}}{4}\\
- \frac{{36b}}{2} + 12 + 2a + \frac{{25b}}{2} - \frac{{39}}{4} = \frac{9}{4} + 2a - \frac{{11b}}{2}\\
\frac{9}{4} + 2a - \frac{{11b}}{2} = 1
\end{array} \right. ⎩ ⎨ ⎧ 4 3 2 − 3 + 1 = 3 a − 3 a + 4 1 5 a − 3 a + 4 1 = − 2 25 b + 10 + 2 a + 2 25 b − 4 39 − 2 36 b + 12 + 2 a + 2 25 b − 4 39 = 4 9 + 2 a − 2 11 b 4 9 + 2 a − 2 11 b = 1
{ 1 4 = 1 4 2 a + 1 4 = 2 a + 1 4 − 11 b 2 + 9 4 + 2 a = 9 4 + 2 a − 11 b 2 2 a − 11 b 2 = − 5 4 \left\{ \begin{array}{l}
\frac{1}{4} = \frac{1}{4}\\
2a + \frac{1}{4} = 2a + \frac{1}{4}\\
- \frac{{11b}}{2} + \frac{9}{4} + 2a = \frac{9}{4} + 2a - \frac{{11b}}{2}\\
2a - \frac{{11b}}{2} = - \frac{5}{4}
\end{array} \right. ⎩ ⎨ ⎧ 4 1 = 4 1 2 a + 4 1 = 2 a + 4 1 − 2 11 b + 4 9 + 2 a = 4 9 + 2 a − 2 11 b 2 a − 2 11 b = − 4 5
2 a − 11 b 2 = − 5 4 ⇒ 8 a − 22 b = − 5 ⇒ a = − 5 + 22 b 8 2a - \frac{{11b}}{2} = - \frac{5}{4} \Rightarrow 8a - 22b = - 5 \Rightarrow a = \frac{{ - 5 + 22b}}{8} 2 a − 2 11 b = − 4 5 ⇒ 8 a − 22 b = − 5 ⇒ a = 8 − 5 + 22 b
This equation has an infinite number of solutions. Let's find any particular solution.
Let b = 5 22 ⇒ a = 0 b = \frac{5}{{22}} \Rightarrow a = 0 b = 22 5 ⇒ a = 0
Then
f ( x ) = { 1 2 ( x − 2 ) , 2 ≤ x ≤ 3 0 , 3 < x ≤ 5 2 − 5 22 x , 5 < x ≤ 6 0 , o t h e r w i s e f(x) = \left\{ {\begin{matrix}
{\frac{1}{2}(x - 2),\,\,2 \le x \le 3}\\
{0,\,\,3 < x \le 5}\\
{2 - \frac{5}{{22}}x,\,\,5 < x \le 6}\\
{0,\,\,otherwise}
\end{matrix}} \right. f ( x ) = ⎩ ⎨ ⎧ 2 1 ( x − 2 ) , 2 ≤ x ≤ 3 0 , 3 < x ≤ 5 2 − 22 5 x , 5 < x ≤ 6 0 , o t h er w i se
F ( x ) = { 0 , x < 2 x 2 4 − x + 1 , 2 ≤ x ≤ 3 0 − 0 + 1 4 , 3 < x ≤ 5 − 5 x 2 44 + 2 x + 125 44 − 39 4 , 5 < x ≤ 6 9 4 − 55 44 , x > 6 = { 0 , x < 2 x 2 4 − x + 1 , 2 ≤ x ≤ 3 0 − 0 + 1 4 , 3 < x ≤ 5 − 5 x 2 44 + 2 x − 76 11 , 5 < x ≤ 6 1 , x > 6 F(x) = \left\{ {\begin{matrix}
{0,\,\,x < 2}\\
{\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\
{0 - 0 + \frac{1}{4},\,\,3 < x \le 5}\\
{ - \frac{{5{x^2}}}{{44}} + 2x + \frac{{125}}{{44}} - \frac{{39}}{4},\,\,5 < x \le 6}\\
{\frac{9}{4} - \frac{{55}}{{44}},\,\,x > 6}
\end{matrix}} \right.= \left\{ {\begin{matrix}
{0,\,\,x < 2}\\
{\frac{{{x^2}}}{4} - x + 1,\,\,2 \le x \le 3}\\
{0 - 0 + \frac{1}{4},\,\,3 < x \le 5}\\
{ - \frac{{5{x^2}}}{{44}} + 2x - \frac{{76}}{{11}},\,\,\,5 < x \le 6}\\
{1,\,\,x > 6}
\end{matrix}} \right. F ( x ) = ⎩ ⎨ ⎧ 0 , x < 2 4 x 2 − x + 1 , 2 ≤ x ≤ 3 0 − 0 + 4 1 , 3 < x ≤ 5 − 44 5 x 2 + 2 x + 44 125 − 4 39 , 5 < x ≤ 6 4 9 − 44 55 , x > 6 = ⎩ ⎨ ⎧ 0 , x < 2 4 x 2 − x + 1 , 2 ≤ x ≤ 3 0 − 0 + 4 1 , 3 < x ≤ 5 − 44 5 x 2 + 2 x − 11 76 , 5 < x ≤ 6 1 , x > 6
sketch the graph of f(x):
sketch the graph of F(x):
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