Try this:Originally Posted by epv
This expresses the fact that 1/5 of the probability is concentrated at x=0, 2/5 is spread out uniformly over [1,3], and 2/5 is concentrated at x=5.
Hello,
I was a bit confused by this problem that appeared in my prob and stat textbook:
"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. Sketch the d.f. of Y"
I wanted to solve this problem by first finding the mixed probability function of Y, f_Y. To do this I first found the probability function of X, f_X:
f_X(x) = { 0 if x < 0; 1/5 if 0 <= x <= 5; 0 if x > 5 } by the fact that X has uniform distribution on [0,5].
Then to find f_Y, the mixed probability function of Y:
Pr(Y = 0) = Pr(X <= 1) = 1/5
Also, it's given that Y = 5 if X >= 3 which means that
Pr(Y = 5) = Pr(X >= 3) = 2/5
The only solution I could think of was that after summing the places of non-zero probability mass at Y = 0 and Y = 5 to get 1/5 + 2/5 = 3/5, that then it forces Pr(0 < Y < 5) = 2/5. Then maybe would I interpret "Y = X otherwise" to mean that simply Y distrubutes the probability 2/5 uniformly over the interval (0,5)? In that case I would have:
f_Y(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 }
And then the Pr(Y = 0) + Pr(0 < Y < 5) + Pr(Y = 5) = 1/5 + 10/25 + 2/5 =
25/25 = 1 as needed. But I'm not sure that that's what is meant by "Y = X otherwise" Oh well, I'd really appreciate any help on determining the correct formulas for the mixed probability function of Y. Thanks in advance for any help,