# Homework Help: Finding CDF for Continuous RVs problem

1. Mar 29, 2012

1. The problem statement, all variables and given/known data
I'm given the pdf and asked to find F(y)/ cdf. I've calculated it many times, but I'm not getting the right numbers. the pdf is

f(y)= .5, .............-2≤y≤0
.75-.25y,.....1≤y≤3
0,................elsewhere

so that means

f(y)= 0,............y< -2
0.5, ..........-2≤y≤0
0,................0<y<1
.75-.25y,.....1≤y≤3
0,................ y>3

right?

So, my CDF is

F(y)= c1,................................ y<-2
.5y +c2,......................... -2≤y≤0
c3, ...................................0<y<1
.75y- .125y^2+ c4,........... 1≤y≤3
c5,......................................y>3

so, I found the constants, but I'm not getting c5=1.

0=c1=.5y+c2 at y=-2
so 1=c2

.5y+1=c3 at y=0
so 1=c3

1=c3=.75y- .125y^2+ c4 at y=1
1= .75- .125+ c4
1-.625= .375=c4

.75y-.125y^2+ .375= c5 at y=3
.75*3 -.125(9) +.375= c5
2.25 - 1.125 + .375= 1.5=c5≠1

I don't know where I'm going wrong. Could you show me?? -_- Much thanks.

I hope this is legible....

2. Mar 29, 2012

### RoshanBBQ

Assuming you've given the right problem statement, I think the question is just flawed. Take a step back from calculus for a bit and consider you are just accumulating area. The function is zero before y = -2, so at that point the accumulation is zero. You then accumulate a rectangle of length 2 (from -2 to 0) and height .5. The area accumulated it 2*.5 = 1. You've already gotten 1 area... so the fact you will be accumulating more is a problem.

3. Mar 29, 2012

@RoshanBBQ,

Yeah, I looked at it from a geometric pov and it came out to be more than 1, but I've worked this problem many times, with a calculator and without... being careful with the numbers and such... I've just e-mailed my professor, so we shall see D:.... Thanks for your quick response!

4. Mar 29, 2012

### Ray Vickson

For any z > -2 the CDF is $F(z) = \int_{-2}^z f(y) \, dy,$ so there is no need to introduce arbitrary constants of integration. However, your f(y) integrates to 3/2 > 1, so is not a legitimate probability density function.

RGV