Question

Let X be a random variable with CDF z<0 G()=/2 0 <IS2 z>2 1 Suppose Y = X2 is another random variable, find (a) P(1/2 X 3/2), (b) P(1s X< 2) (c) P(Y X) (d) P(X 2Y). (f) If Z VX, find the CDF of Z. (d) P(X+Y 3/4)

Answer 1

We would be looking at the first 4 parts here as:

a) P(0.5 < X < 1.5)
= F(1.5) - F(0.5)

= (1.5/2) - (0.5/2) = 0.5

Therefore 0.5 is the required probability here.

b) The probability here is computed as:
P(1 <= X < 2)

= F(2) - F(1)

= 1 - (1/2)

= 0.5

Therefore 0.5 is the required probability here.

c) The PDF for X here is first obtained here by differentiating the CDF with respect to x as:

f(x) = 1/2, 0 < x < 2

The probability now is computed here as:

PY <X) = P(X? – X<0)

As we know here that X > 0, therefore X < 1 for the above equation to be true here.

PY <X) = P(X? – X<0) = P(X<1) = F(1) = 0.5

Therefore 0.5 is the required probability here.

d) The probability here is computed here as:

P(X <2Y) = P(X < 2x) = P(X(2X – 1) > 0) = P(2X – 1 > 0

= P(x > 0.5) = 1 – F(0.5) = 1 - 0,5 = 0.75

Therefore 0.75 is the required probability here.