# Uniform distribution- probabilities

1. Mar 21, 2015

### Dassinia

Hello, I am stuck at this exercise:

1. The problem statement, all variables and given/known data

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

2. Relevant equations
3. The attempt at a solution

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

2. Mar 21, 2015

### haruspex

I'm not familiar with the notation min(X; a=2). What does it mean?

3. Mar 21, 2015

### Ray Vickson

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 $a > 2$.

4. Mar 21, 2015

### Dassinia

Oh sorry, I didn't notice that there is a mistake in the expression of Y

It is Y=min(X; a/2)

5. Mar 21, 2015

### haruspex

In that case your CDF for Y is correct. Draw it. Is it continuous?

6. Mar 21, 2015

### Dassinia

No, it is not continuous !
But how to find the cumulative distribution function ?
Thanks !

7. Mar 21, 2015

### haruspex

The CDF is what you wrote in the OP. You specified P(Y<=t) for all three ranges of t.

8. Mar 21, 2015

### Dassinia

Oh, right ! I don't know why I thought that I had to find something else