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