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Let x1 x2 x3 be three independent and identically distribuyed random variables with uniform distrbution on (0,1) find probability p(x1+x2<=x3)
asked in Probability by Veteran (31.1k points)   | 464 views

3 Answers

0 votes
Best answer

http://math.stackexchange.com/questions/688174/uniform-distribution-in-0-1-where-px1x2-x3

Refer to the above link,solution is nicely explained..

 

 

 

 

answered by Loyal (3.5k points)  
+2 votes

The possible cases are :

X1+X2<=X3

X1+X3<=X2

X2+X3<=X1

X2+X1>=X3

X2+X3>=X1

X3+X1>=X2

Out of these 6 possible cases only 1 is favorable . Therefore Ans is  1/6 .

answered by Boss (6.9k points)  
the cases you have stated are not mutually exclusive..
0 votes

I guess the right answer should be 0.25

(x1 + x2) is a new random variable that can take any values in the interval (0, 2) (that is (0+0, 1+1)).

(x1 + x2) will be uniformly distributed over (0, 2) since x1 & x2 are independent and are uniformly distributed over their respective domains.

Since (x1 + x2) is uniformly distributed over its domain,

P((x1 + x2) is greater than or equal to 1) = P((x1 + x2) is less than 1) = 0.5

Now consider the following two cases:

Case 1:

When (x1 + x2) will take any value greater than or equal to 1, no matter what the value of x3 is, it will always be greater than x3.

That is, P((x1 + x2) is greater than x3 given that (x1+x2) is greater than or equal to 1) = 1.

Case 2:

When (x1 + x2 ) will take any value less than 1, it may or may not be greater than x3.

Since random variables x1, x2 & x3 are independent of each other, (x1+x2) will also be independent of x3.

Now since (x1 + x2) & x3 are independent, any of them can be greater than the other, when (x1 + x2) can take values between 0 & 1 only.

That is, P((x1 + x2) is greater than x3 given that (x1 + x2) is less than 1) = 0.5


Summing up the above cases using conditional probabilities, we can say that, 

P((x1 + x2) is greater than x3) = {P((x1 + x2) is greater than or equal to 1)*P((x1 + x2) is greater than x3 given that it is greater than  or equal to 1)} + {P((x1 +x2) is less than 1)*P((x1 + x2) is greater than x3 given that it is less than 1)}

P((x1 + x2) is greater than x3) = {0.5 * 1} + {0.5 * 0.5} = 0.75

Now,

P((X1 + x2) is less than or equal to x3) = 1 - P((x1 + x2) is greater than x3) = 1 - 0.75 = 0.25

 

 

 

answered by Veteran (12.8k points)  
edited by


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