Question
A uniformly distributed random variable X with probability density function
fX(x) = (1/10) ( u(x+5) − u(x−5) )
where u(.) is the unit step function, is passed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be...
Y =
1
when X ∈ [−2.5, 2.5], else 0
(a) fY(y) = (1/5)(u(y+2.5) − u(y−2.5))
(b) fY(y) = 0.5δ(y) + 0.5δ(y−1)
(c) fY(y) = 0.25δ(y+2.5) + 0.25δ(y−2.5) + 0.5δ(y)
(d) fY(y) = 0.25δ(y+2.5) + 0.25δ(y−2.5)
Correct Answer
The transformation maps X to Y such that:
- If X ∈ [-2.5, 2.5], then Y = 1
- If X < −2.5 or X > 2.5, then Y = 0
Since X is uniform on [-5, 5], the probability that X is inside [-2.5, 2.5] is:
P(Y=1) = 5/10 = 0.5
P(Y=0) = 0.5
P(Y=0) = 0.5
Thus the PDF of Y is:
fY(y) = 0.5δ(y) + 0.5δ(y − 1)
* This corresponds to option (b)