1. ## CDF question

I am a bit confused here and really have to understand where I am lost.

Suppose the CDF $F(x) = 0 \ \ for \ \ x <1;\ \ \dfrac{x^2-2x+2}{2}\ \ for \ \ 1 \leqq x<2 \ \ and \ 1 \ \ for \ \ x \geqq 2$

So if x<1 we have F(x)=0. That is, the sum of all probabilities from - $\infty$ up to but not including 1 is 0. But F(1)=1. So I am concluding that f(1)=1 (f(x) is the pdf). So far I think I am ok and understanding things.

Now, given F(x), I claim f(x)= x-1 for 1 $\leqq$x<2 and 0 elsewhere. Still seems good to me until....

Using my formula for f(x) I get f(1)=0 which contradicts what I got above. Also the integral from 1 to 2 of f(x)dx is not 1. What is going on here?

I am sure that what I am missing here is very important so I hope that someone can please carefully explain this to me.

Thank you!

2. ## Re: CDF question

Originally Posted by JaguarXJS
Suppose the CDF $F(x) = 0 \ \ for \ \ x <1;\ \ \dfrac{x^2-2x+2}{2}\ \ for \ \ 1 \leqq x<2 \ \ and \ 1 \ \ for \ \ x \geqq 2$

Now, given F(x), I claim f(x)= x-1 for 1 $\leqq$x<2 and 0 elsewhere. Still seems good to me until....CORRECT!

Using my formula for f(x) I get f(1)=0 which contradicts what I got above. Also the integral from 1 to 2 of f(x)dx is not 1. What is going on here? FALSE, FALSE!

CDF: $F(x)=\begin{cases}0 &, x< 1 \\\displaystyle\int_1^x {(t - 1)dt} &, 1\le x<2\\1 &, 1\le x \end{cases}$

3. ## Re: CDF question

Hold on (I'm sure you are correct but I'm not..yet)- The integral of (t-1)dt = {(t-1)^2}/2 from 1 to x and that is {(X-1)^2} = (x^2-2x+1)/2 NOT (x^2-2x+2)/2.

Also is this statement true?: So if x<1 we have F(x)=0. That is, the sum of all probabilities from -infinity up to but not including 1 is 0. But F(1)=1. So I am concluding that f(1)=1 (f(x) is the pdf).

4. ## Re: CDF question

Originally Posted by JaguarXJS
Hold on (I'm sure you are correct but I'm not..yet)- The integral of (t-1)dt = {(t-1)^2}/2 from 1 to x and that is {(X-1)^2} = (x^2-2x+1)/2 NOT (x^2-2x+1)/2.

Also is this statement true?: So if x<1 we have F(x)=0. That is, the sum of all probabilities from -infinity up to but not including 1 is 0. But F(1)=1. So I am concluding that f(1)=1 (f(x) is the pdf).
$\displaystyle{\int_1^x {(t - 1)dt}} = \left. {\left( {\frac{{{t^2}}}{2} - t} \right)} \right|_1^x = \left( {\frac{{{x^2}}}{2} - x} \right) - \left( {\frac{{{1^2}}}{2} - 1} \right) = \frac{{{x^2} - 2x + 2}}{2}$

Learn to do calculus otherwise there is no point in your trying to do this material.

5. ## Re: CDF question

Excuse me but 1/2 - 1 is not -1 but rather -1/2.

As I am sure that you know, making arithmetic mistakes does not mean you can't be a top mathematician as you obviously are.

Also, you can say that int (t-1)dt = [(t-1)^2]/2.

6. ## Re: CDF question

Originally Posted by JaguarXJS
Excuse me but 1/2 - 1 is not -1 but rather -1/2.

As I am sure that you know, making arithmetic mistakes does not mean you can't be a top mathematician as you obviously are.

Also, you can say that int (t-1)dt = [(t-1)^2]/2.
You cannot even do basic arithmetic!
$- \left( {\frac{{{1^2}}}{2} - 1} \right) = - \left( { - \frac{1}{2}} \right) = + \frac{1}{2}$

7. ## Re: CDF question

I knew that I did not put the neg sign in front!

All I am asking you to do is PLEASE look at YOUR own (correct) work and see that YOU got 2/2, not 1/2

You got $\dfrac {x^2 - 2x + 2}{2} NOT \dfrac {x^2 - 2x + 1}{2}.$ After all 2/2 is not 1/2

8. ## Re: CDF question

Originally Posted by Plato
$\displaystyle{\int_1^x {(t - 1)dt}} = \left. {\left( {\frac{{{t^2}}}{2} - t} \right)} \right|_1^x = \left( {\frac{{{x^2}}}{2} - x} \right) - \left( {\frac{{{1^2}}}{2} - 1} \right) = \frac{{{x^2} - 2x + 2}}{2}$

Learn to do calculus otherwise there is no point in your trying to do this material.
$-(\dfrac {1^2}{2} - 1)$ is 1/2 but please LOOK as YOU got 2/2.