SOA 123 Prob 62

[The_Czar52]

A r.v. X has the CDF:

F(x) = {0, x<1
(x^2-2x+2)/2, 1<=x<2,
1, x>2
}

Calculate the variance of X.

The solution immediately says "note that the density function of X is..." but doesn't go on to say how we come up with it. I guess I'm just confused on mixed distributions. How do we come up with the PDF for a mixed distribution from a given CDF? Calculating the variance is straightforward once you have the PDF but I just can't grasp how to get there.

When I differentiate the CDF I get the correct PDF but I'm off by a constant. How do we know exactly how to do this? What if the "weights" of each are something different from 1/2?

Could someone help me out and give a full explanation of how they would approach/solve this problem. Thanks a lot and I appreciate any help.

http://www.casact.org/admissions/studytools/exam1/P-09-07QS.pdf

[jthias]

The trick is to look at the endpoints of the pieces of the piecewise-defined F.

F(1) = 1/2 and F(x) = 0 for x<1 forces f(1) to be 1/2.

lim as x --> 2- of F(x) is 1, so X has no non-zero probability at the point x=2.

So to conclude, X is a mixed distribution with

f(1) = 1/2

and

f(x) = x-1 on (1,2)

and zero elsewhere

my apologies for the mistakes, I have bad habit of working out problems via keyboard sometimes..turns out I had it right the first time around :(

[ctperng]

A r.v. X has the CDF:

F(x) = {0, x<1
(x^2-2x+2)/2, 1<=x<2,
1, x>2
}

Calculate the variance of X.

The solution immediately says "note that the density function of X is..." but doesn't go on to say how we come up with it. I guess I'm just confused on mixed distributions. How do we come up with the PDF for a mixed distribution from a given CDF? Calculating the variance is straightforward once you have the PDF but I just can't grasp how to get there.

When I differentiate the CDF I get the correct PDF but I'm off by a constant. How do we know exactly how to do this? What if the "weights" of each are something different from 1/2?

Could someone help me out and give a full explanation of how they would approach/solve this problem. Thanks a lot and I appreciate any help.

http://www.casact.org/admissions/studytools/exam1/P-09-07QS.pdf

As long as you observe it's a mixed distribution and have attached correct weights to the discrete parts, the off-by-constant is not an issue.

To avoid differentiation, you can see that X is a positive random variable, therefore, you can compute the variance by the survival method:

Var(X) = E(X^2) - (E(X))^2
= int(0 to infinity) of 2x*S(x) dx -{int(0 to infinity) of S(x) dx}^2

Please double check, as I didn't bother to check it. :)

ctperng

[Jolin]

I would say, the key word is the "<=" that stated at the question...Everytime you encounter those <= just be extra careful...to be safe check the F(x) before you do the solution...I know this is tricky...i got caught too...:Waldo:

[JGET]

A r.v. X has the CDF:

F(x) = {0, x<1
(x^2-2x+2)/2, 1<=x<2,
1, x>2
}

Calculate the variance of X.

The solution immediately says "note that the density function of X is..." but doesn't go on to say how we come up with it. I guess I'm just confused on mixed distributions. How do we come up with the PDF for a mixed distribution from a given CDF? Calculating the variance is straightforward once you have the PDF but I just can't grasp how to get there.

When I differentiate the CDF I get the correct PDF but I'm off by a constant. How do we know exactly how to do this? What if the "weights" of each are something different from 1/2?

Could someone help me out and give a full explanation of how they would approach/solve this problem. Thanks a lot and I appreciate any help.

http://www.casact.org/admissions/studytools/exam1/P-09-07QS.pdf

I had a bit of a problem figuring this one out too, but the trick is to seprate the
(x^2-2x+2)/2 into 2 parts

0.5*(x^2-2x)+0.5*(2)

Note its actually like saying c*(f(x)) + (1-c)*(g(y)) where c is an arbitrary constant.

From this you should be able to concluded its a mixed distribution.
The endpoints as jthias described, give you an idea of where the point densities exist.

JGET

[The_Czar52]

For some reason I'm still confused on this. When I split it up, it get:

F(x) = .5(x^2-2x+1) + .5(1) = .5(x^2-2x+2) (The original CDF)

F_x1(x) = x^2-2x+1, 1<=x<2,
= 0, x<1

F_x2(x) = 1, x = 1,
= 0, x<1

f_x1(x) = 2x-2, 1<=x<2
f_x2(x) = 1, x=1

Var(X) = E(X^2) - [E(X)]^2

E(X^2) = .5E(X_1^2) + .5E(X_2^2) = .5(17/6) + .5(1)
E(X) = .5E(X_1) +.5E(X_2) =.5(5/3) + .5(1)

Var(X) = 23/12 - (4/3)2 = 5/36 ~ .139

I just worked this out and I actually got the correct answer! (I thought going in that there was no way it would work out). I tried this before and for some reason it never worked. It must have been a stupid algebra mistake everytime that I overlooked.

I find it amazing that we still all got the same answer yet we used 3 or 4 different methods to do it. Math is truely amazing! :geek:

Thanks for the help everyone!

[jthias]

...Math is truely amazing! :geek:

Yup...one of the few things that keeps me awestruck as an adult, the same way I felt as a kid watching a magic act, and wondering how the magician pulled a rabbit from his hat, but math goes one step better...you can show how the magician produced that rabbit :)

[Anu Dhanuka]

For some reason I'm still confused on this. When I split it up, it get:

F(x) = .5(x^2-2x+1) + .5(1) = .5(x^2-2x+2) (The original CDF)

F_x1(x) = x^2-2x+1, 1<=x<2,
= 0, x<1

F_x2(x) = 1, x = 1,
= 0, x<1

f_x1(x) = 2x-2, 1<=x<2
f_x2(x) = 1, x=1

Var(X) = E(X^2) - [E(X)]^2

E(X^2) = .5E(X_1^2) + .5E(X_2^2) = .5(17/6) + .5(1)
E(X) = .5E(X_1) +.5E(X_2) =.5(5/3) + .5(1)

Var(X) = 23/12 - (4/3)2 = 5/36 ~ .139

I just worked this out and I actually got the correct answer! (I thought going in that there was no way it would work out). I tried this before and for some reason it never worked. It must have been a stupid algebra mistake everytime that I overlooked.

I find it amazing that we still all got the same answer yet we used 3 or 4 different methods to do it. Math is truely amazing! :geek:

Thanks for the help everyone!

Hi Czar,
Thats so tru, that same problem in maths have different approaches....i guess this mixed distribution is still '''''ing me up....
I saw this question in ASM....i understood the ways what other guys followed....i am unsure about ur way, and wants to learn this....can u please help me in giving an explaination hw u solved it (step by step preferable)...
plus can u also help me to knw, hw do u break main component of F(x) in its components....
from there after we can then easily get f(x)....

all explaination will be highly appreciated...

Thanks a lot...

[The_Czar52]

Hi Czar,
Thats so tru, that same problem in maths have different approaches....i guess this mixed distribution is still '''''ing me up....
I saw this question in ASM....i understood the ways what other guys followed....i am unsure about ur way, and wants to learn this....can u please help me in giving an explaination hw u solved it (step by step preferable)...
plus can u also help me to knw, hw do u break main component of F(x) in its components....
from there after we can then easily get f(x)....

all explaination will be highly appreciated...

Thanks a lot...

Sure,

The original CDF is .5(x^2-2x+2). I rewrote it as...
.5(x^2-2x+1+1) = .5(x^2-2x+1) + .5(1). This is of the form
c1(F_x1(x) + c2(F_x2(x) where c1,c2 are the weights and F_x1(x) and F_x2(x) are the continuous and discrete components of the CDF.

Now F_x1(x) = x^2-2x+1, 1<=x<2,
= 0, x<1

F_x2(x) = 1, x = 1,
= 0, x<1

Differentiating the continous CDF gives us the continuous PDF, f_x1(x)

f_x1(x) = 2x-2, 1<=x<2

And the discrete component equals 1 at x=1 and 0 elsewhere.
f_x2(x) = 1, x=1

Var(X) = E(X^2) - [E(X)]^2

To find the variance now we simply take the weighted average of E(X^2) and the weighted E(X)

E(X^2) = .5E(X_1^2) + .5E(X_2^2) = .5(17/6) + .5(1) = 23/12
E(X) = .5E(X_1) +.5E(X_2) =.5(5/3) + .5(1) = 4/3

Var(X) = 23/12 - (4/3)2 = 5/36 ~ .139

Hope this helps!

[Anu Dhanuka]

Sure,

The original CDF is .5(x^2-2x+2). I rewrote it as...
.5(x^2-2x+1+1) = .5(x^2-2x+1) + .5(1). This is of the form
c1(F_x1(x) + c2(F_x2(x) where c1,c2 are the weights and F_x1(x) and F_x2(x) are the continuous and discrete components of the CDF.

Now F_x1(x) = x^2-2x+1, 1<=x<2,
= 0, x<1

F_x2(x) = 1, x = 1,
= 0, x<1

Differentiating the continous CDF gives us the continuous PDF, f_x1(x)

f_x1(x) = 2x-2, 1<=x<2

And the discrete component equals 1 at x=1 and 0 elsewhere.
f_x2(x) = 1, x=1

Var(X) = E(X^2) - [E(X)]^2

To find the variance now we simply take the weighted average of E(X^2) and the weighted E(X)

E(X^2) = .5E(X_1^2) + .5E(X_2^2) = .5(17/6) + .5(1) = 23/12
E(X) = .5E(X_1) +.5E(X_2) =.5(5/3) + .5(1) = 4/3

Var(X) = 23/12 - (4/3)2 = 5/36 ~ .139

Hope this helps!

Yeah it did, thanks a lot...but i still have one question in my mind...how did u decide the limits of CDF components? Also according to me the limit of F1(x) shuld be 1<x<2...
Correct me if i am wrong....
Thanks again.

[The_Czar52]

Yeah it did, thanks a lot...but i still have one question in my mind...how did u decide the limits of CDF components? Also according to me the limit of F1(x) shuld be 1<x<2...
Correct me if i am wrong....
Thanks again.

Well for the continuous CDF it doesn't actually start until x=1 because there is a "jump" in the original CDF and if you plug in 0 you get 1 which wouldn't make any sense. The discrete CDF must be zero less than 1 and 1 at 1 since all of the probability is in the point mass at 1. And yes the bounds for F_x1(x) should be 1<x<2 not 1<=x<2 because there is a point mass for the discrete component at that point. Minor but technically correct. Thanks for pointing it out as good actuaries pay attention to detail!