Sum of two uniforms (dependent)

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The discussion focuses on calculating the probability P(X+Y<=z) where X is a uniform random variable and Y is dependent on X through a constant c. The initial attempt at a solution uses the cumulative distribution function for the uniform distribution, but it is noted that this is only valid within the bounds of the uniform distribution, specifically for x in [a, b]. The condition for the probability to hold is established as 2a+c ≤ z ≤ 2b+c. If z falls outside this range, the probability must be adjusted to either 1 or 0. The conversation emphasizes the importance of ensuring that z meets the necessary conditions for the probability calculation to be valid.
steveholy
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Thanks in advance for any feedback.

Homework Statement



X ~ U(a,b) ( X is a Uniform random variable within the interval a-b.)
Y= X+constant

Say, constant = c.

P(X+Y<= z) = ?

Homework Equations



P(X<=x) = (x- a) / (b-a) (The cumulative distribution of a Uniform R.V.)

The Attempt at a Solution



P(X+Y<=z) = P(X+X+c<=z) = P(X<= (z-c)/2) =( (z-c) /2 - a) / (b-a).

Y is dependent on X, that got me confused in being 100% sure that my solution above is correct.

Thanks.
 
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Hi steveholy! :smile:

The general method looks ok. But:

steveholy said:
P(X<=x) = (x- a) / (b-a) (The cumulative distribution of a Uniform R.V.)

This is only true for x\in [a,b]. Hence

P(X<= (z-c)/2) =( (z-c) /2 - a) / (b-a).

this is only true for

a\leq \frac{z-c}{2}\leq b

thus if

2a+c\leq z\leq 2b+c

if z does not satisfy this, then you'll need to write that the probability is 1 or 0 (depending on the value of z).
 
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