- #1
kuahji
- 394
- 2
Find the distribution function of the random variable X whose probability density is given by
f(x)= x/2 for x < x [tex]\leq1[/tex]
1/2 for 1< x [tex]\leq2[/tex]
(3-x)/2 for 2 < x [tex]\leq3[/tex]
0 elsewhere
Ok, so I know have to take the integral of each piece, but I'm having a hard time figuring out the limits of the integrals.
The book has the answer
F(x)= 0 for x [tex]\leq0[/tex]
x^2/4 for 0 < x [tex]\leq1[/tex]
1/4(2x-1) 1 < x [tex]\leq2[/tex]
1/4(6x-x^2-5) for 2 < x < 3
1 for x[tex]\geq3[/tex]
Which is about what I got when I took the "indefinite" integrals of each piece (again, not sure how to go about setting the limits), but for example in this section 1/4(6x-x^2-5) for 2 < x < 3 where is the -5 coming from? Any help would be appreciated.
f(x)= x/2 for x < x [tex]\leq1[/tex]
1/2 for 1< x [tex]\leq2[/tex]
(3-x)/2 for 2 < x [tex]\leq3[/tex]
0 elsewhere
Ok, so I know have to take the integral of each piece, but I'm having a hard time figuring out the limits of the integrals.
The book has the answer
F(x)= 0 for x [tex]\leq0[/tex]
x^2/4 for 0 < x [tex]\leq1[/tex]
1/4(2x-1) 1 < x [tex]\leq2[/tex]
1/4(6x-x^2-5) for 2 < x < 3
1 for x[tex]\geq3[/tex]
Which is about what I got when I took the "indefinite" integrals of each piece (again, not sure how to go about setting the limits), but for example in this section 1/4(6x-x^2-5) for 2 < x < 3 where is the -5 coming from? Any help would be appreciated.