Uniform distribution- probabilities

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Dassinia
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Hello, I am stuck at this exercise:

1. Homework Statement

X ~ U(0, a), a > 0 and Y = min(X; a=2).
- Find the cumulative distribution function of Y
-Is the variable Y continuous ?

Homework Equations


3. The Attempt at a Solution [/B]
The density function for X is
f(t)= 1/a if 0≤t≤a , 0 elsewhere
Is it correct to write that :
∀t < 0, P(Y ≤ t) = 0
∀0 ≤ t < a/2, P(Y ≤ t) = P(X ≤ t) = t/a
∀t ≥ a/2, P(Y ≤ t) = 1

And then don't know what to do
Thanks
 
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Dassinia said:
Hello, I am stuck at this exercise:

1. Homework Statement

X ~ U(0, a), a > 0 and Y = min(X; a=2).
- Find the cumulative distribution function of Y
-Is the variable Y continuous ?

Homework Equations


3. The Attempt at a Solution [/B]
The density function for X is
f(t)= 1/a if 0≤t≤a , 0 elsewhere
Is it correct to write that :
∀t < 0, P(Y ≤ t) = 0
∀0 ≤ t < a/2, P(Y ≤ t) = P(X ≤ t) = t/a
∀t ≥ a/2, P(Y ≤ t) = 1

And then don't know what to do
Thanks

You have an '##a##' in the definition of ##X## itself, and an '##a##' in the "definition" of ##Y## in terms of ##X##. Are they supposed to be the same '##a##' in both places? If so, I cannot make any sense out of what you wrote.

On the other hand, if you really mean that ##Y = \min(X,2)##, then that would have meaning. In that case it is important to distinguish between the two cases ##0 < a \leq 2## and ##a > 2##.
 
Oh sorry, I didn't notice that there is a mistake in the expression of Y

It is Y=min(X; a/2)
 
No, it is not continuous !
But how to find the cumulative distribution function ?
Thanks !
 
Oh, right ! I don't know why I thought that I had to find something else :eek:
Thanks for your answers !