I was not sure whether my work for this problem is correct and I'm more pressed for time:

"Suppose that X had a uniform distribution on the interval [0,5] and that the random variable Y is defined by Y = 0 if X <= 1, Y = 5 if X >= 3, and

Y = X otherwise. Find the mixed probability function of Y"

First I found:

Pr(Y=0)=Pr(X<=1)=1/5

and

Pr(Y=5)=Pr(X>=3)=2/5

because of X's uniform distribution on [0,5]

Then for the rest of the interval [0,5] / {0,5} = (0,5), this forces

Pr(0<Y<5) = 1 - Pr(Y=0) + Pr(Y=5) = 1 - 1/5 - 2/5 = 2/5 = Pr(1<=X<=3).

So, this is where I am not so certain. For 0<Y<5, the probability of 2/5 must be distributed over this interval, and it must be done so uniformly because when 0<Y<5, Y=X and X is distributed uniformly (?).

So, then, if f is the (mixed) probability function of Y, then I would define

f(y) = { 0 if y < 0; 1/5 if y = 0; 2/25 if 0 < y < 5; 2/5 if y = 5; 0 if y > 5} ?

Because then I would have

Pr(Y=0)+Pr(0<Y<5)+Pr(y=5) = 1/5 + Integral[ 2/25 dt,{t, 0, 5} ] + 2/5 =

5/25 + 10/25 + 10/25 = 1 as needed.

So, what I'd like help with is if this answer for the probability function, f, of Y is correct, and if it is not, where my reasoning went wrong.

Thanks for anyone's help in advance,

E