# where am i going wrong - sum of two uniform variates.

• Jun 24th 2010, 08:43 AM
Pushpa
where am i going wrong - sum of two uniform variates.
Problem : Let
X have a uniform distribution on the interval (1; 3).What is the proba-

bility that the sum of 2 independent observations of
X is greater than 5?
Solutions :

Let X1 and X2 be the two independent obeservations.
We have to show P(X1+X2 > 5)=1- P(X1+X2 <=5).

Here f(x1,x2)=1/2 for 1<x1,x2<3
= 0 otherwise
(Is the joint pmf and its range correct??? )
here x1 varies from 1 to 3 and x2 varies from 1 to 5-x1.
Now P(x1+x2<=5)= integral (1<x1<3) integral (1<x2< 5-x1) of 1/2 dxdy

on solving this i am getting answer 2....which is wrong( as probablity can not be more than 1)
the right answer of P(X1+X2 >5)=1/8...
Where i am going wrong please let me know...
thanks....

• Jun 24th 2010, 08:58 AM
SpringFan25
the pdf (not pmf) isn't correct.

for indepdant variables
f(x,y)=f(x)f(y) = 0.5*0.5 = 0.25

I'm not sure the limits on your integrals are right either. you cant, for example have x1=1 and x2=(5-1). You only integrate over values of x1,x2 that are acheivable.
• Jun 24th 2010, 02:47 PM
theodds
Springfan is correct on both counts. The pdf is given by

$\displaystyle f(x_1, x_2) = \begin{cases}\frac{1}{4} & \qquad 1 \le x_1 \le 3, 1 \le x_2 \le 3 \\ 0 & \qquad o/w \end{cases}$.

The relevant integral becomes

$\displaystyle \int_1 ^ 3 \int_1 ^ {5 - x_1} f(x_1, x_2) \ dx_1 dx_2$

but remember that this integral isn't directly telling you where the pdf is 0, i.e. if you try to integrate 1/4 over this area you will get the wrong answer. You want to integrate 1/4 over the set such that BOTH $\displaystyle x_1, x_2 \in [1, 3]$ and $\displaystyle x_1 \le 5 - x_2$.

I did the problem and got $\displaystyle P(X_1 + X_2 \le 5) = \frac{7}{8}$.
• Jun 25th 2010, 08:38 AM
Pushpa
I got it.....
Thanks guys.....
I really appreciate you help....
• Jun 25th 2010, 11:15 PM
matheagle
you can obtain the probability quicker with geometry than calculus