can someone explain this probability solution?

Question

let X have pdf f(x)=4x^3, 0
the answer in the back is g(y)= 2y, 0
im not sure how they got this answer, thanks

the question i need to answer is let X have pdf f(x) = xe^(-x^2/2), o
is it the similiar process as the question above?

Answer 1

First, find the cdf F_X(x):

F_X(x) = integral from 0 to x of f(x)dx = x^4, 0 < x < 1
F_X(x) = 1 for x ≥ 1 and F_X(x) = 0 for x ≤ 0

Now, F_X(x) = P(X < x)

By definition of the cdf G_Y(y) for Y = X²,

G_Y(y) = P(Y < y)
G_Y(y) = P(X² < y)

Therefore, G_Y(y) = 0 if y ≤ 0 (since X² ≥ 0); otherwise,

G_Y(y) = P(-√y < X < √y)
G_Y(y) = P(X < √y) - P(X < -√y)
G_Y(y) = F_X(√y) - F_X(-√y)
G_Y(y) = F_X(√y) (since F_X(a) = 0 for a < 0)

Therefore, for y>0,

G_Y(y) = F_X(√y) = (√y)^4 for 0 < √y < 1, and
G_Y(y) = 1 for √y ≥ 1

That is,

G_Y(y) = y² for 0 < y < 1, and G_Y(y) = 1 for y ≥ 1

So, in sum,

G_Y(y) = 0 for y ≤ 0,
G_Y(y) = y² for 0 < y < 1, and
G_Y(y) = 1 for y ≥ 1

Then g(y) = dG/dy = 2y for 0 < y < 1 and 0 otherwise.

Answer 2

The probability that Y P[Y
Integral[0,y] pdf(Y) dY= Integral[0,y^(1/2)] pdf(x)dx

Take derivative of both sides to find: pdfY(y) = pdfX(y^(1/2))/2y^(1/2) = 4(y^(1/2)^3/2y^(1/2)=2 y.

In your notation, g(y) = 2y

Answer 3

my answer is 11.