# Cumulative Frequency

• Jul 6th 2010, 02:54 AM
cyt91
Cumulative Frequency
A random variable X is uniformly distributed between 0 and 1. Two independent observations are made,X1 and X2. Take (X1,X2 ) as a point on the lines X1 +X2 =Y in a cartesian plane. X1 +X2 =Y is triangular.
(a) show that , for 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= ½ Y^2

(b) show that , for 1≤ Y≤ 2, P( X1 +X2 ≤ Y)=1- ½ (2-Y)^2

I know that f(x)=1 for 0≤ x≤ 1 since X is uniformly distributed. But how do I solve (a)?
Can anyone show me the solution for (a) only so that I could solve (b) myself?

Thanks a lot!(Hi)
• Jul 6th 2010, 07:06 AM
mr fantastic
Quote:

Originally Posted by cyt91
A random variable X is uniformly distributed between 0 and 1. Two independent observations are made,X1 and X2. Take (X1,X2 ) as a point on the lines X1 +X2 =Y in a cartesian plane. X1 +X2 =Y is triangular.
(a) show that , for 0≤ Y≤ 1, P( X1 +X2 =Y)= ½ Y2

(b) show that , for 1≤ Y≤ 2, P( X1 +X2 =Y)=1- ½ (2-Y)2

I know that f(x)=1 for 0≤ x≤ 1 since X is uniformly distributed. But how do I solve (a)?
Can anyone show me the solution for (a) only so that I could solve (b) myself?

Thanks a lot!(Hi)

Is Y defined to be the random variable X1 + X2? What is Y2? P( X1 +X2 =Y) makes no sense, you cannot have the probability of a random variable being equal to another random variable.

Please post the question exactly as it is writen from wherever you got it from.

Note: Yes, the random variable Y = X1 + X2 does follow a triangular distribution. Its pdf is equal to y if $0 \leq y \leq 1$, 2 - y if $1 \leq y \leq 2$ and zero otherwise.
• Jul 6th 2010, 02:19 PM
cyt91
I've edited the question. Sorry for the typo.
How do we know that pdf of Y=X1+X2 is equal to y if 0≤ y≤ 1, 2 - y if 1≤ y≤ 2 and zero otherwise?
• Jul 6th 2010, 03:29 PM
mr fantastic
Quote:

Originally Posted by cyt91
I've edited the question. Sorry for the typo.
How do we know that pdf of Y=X1+X2 is equal to y if 0≤ y≤ 1, 2 - y if 1≤ y≤ 2 and zero otherwise?

1. Y denotes a random variable. So P( X1 +X2 ≤ Y)=1- ½ (2-Y)^2 still makes no sense to me for the reason previously given.

2. Find and read the relevant example in here: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
• Jul 7th 2010, 01:17 AM
cyt91
Well...that is the original question...
How bout this question?

The random variable X is uniformly distributed over the interval [0,9]. Find the probability density function of Y where Y=X^0.5.

The solution given is as follows:

f(x)= {1/9 , 0≤ x≤9
0, otherwise

Let g(y) be the pdf of Y.

The random variable Y is given by Y=X^0.5

G(y)= P( Y ≤ y)
= P( X^0.5 ≤ y)
= P( X ≤ y^2)
= $\int_{0}^{y^2}~f(x)~dx$
= $\int_{0}^{y^2}~1/9~dx$
=(1/9)(y^2)

They defined P( X^0.5 ≤ y). So why not P( X1 +X2 ≤ Y)?

Thanks for the book btw.
• Jul 7th 2010, 04:52 AM
mr fantastic
Quote:

Originally Posted by cyt91
Well...that is the original question...
How bout this question?

The random variable X is uniformly distributed over the interval [0,9]. Find the probability density function of Y where Y=X^0.5.

The solution given is as follows:

f(x)= {1/9 , 0≤ x≤9
0, otherwise

Let g(y) be the pdf of Y.

The random variable Y is given by Y=X^0.5

G(y)= P( Y ≤ y)
= P( X^0.5 ≤ y)
= P( X ≤ y^2)
= $\int_{0}^{y^2}~f(x)~dx$
= $\int_{0}^{y^2}~1/9~dx$
=(1/9)(y^2)

They defined P( X^0.5 ≤ y). So why not P( X1 +X2 ≤ Y)?

Thanks for the book btw.

Do you understand the difference in meaning between Y and y ....?
Quote:

0≤ Y≤ 1, P( X1 +X2 ≤ Y)= ½ Y^2
is NOT the same as 0≤ y≤ 1, P( X1 +X2 ≤ y)= ½ y^2.

Correct notation is important. Incorrect use of notation leads to confusion.
• Jul 8th 2010, 01:11 AM
cyt91
Well, that is the original question. It is given as 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= ½ Y^2 (capital Y).
Anyway,what is the difference between 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= ½ Y^2 and 0≤ y≤ 1, P( X1 +X2 ≤ y)= ½ y^2.
I'm doing self-study and I'm pretty confused with this chapter....