The Probability Density of X^2?

[Kior]

Here is a question about probability density. I am trying to work it out using a different method from the method on the textbook. But I get a different answer unfortunately. Can anyone help me out?

Question:
Let X be uniformly distributed random variable in the internal [ 0, 1]. Find the probability density of X^2?

My trial:
FY(y) = P(Y≤y) = P(X^2≤y) = P(X≤√y) = FX(√y) = ∫ (from 0 to √y) t dt = 0.5 y ⟹ 0.5 = fY(y).
This is actually inspired by http://math.stackexchange.com/questions/...

Solution on the textbook:
y = x^2
dy = 2x dx
h(y)dy = 1 dx
h(y) 2x dx = dx
h(y) = 0.5/x = 0.5/√y

 

[haruspex]

Why t dt, not just dt? And your long line of equalities is bracketed with F_Y(y)=...=f_Y(y), which is clearly not true. Have you missed a step in typing it out?

 

[Kior]

Why t dt, not just dt? And your long line of equalities is bracketed with F_Y(y)=...=f_Y(y), which is clearly not true. Have you missed a step in typing it out?

Thanks get it

Density function should be 1

 

[SammyS]

... And your long line of equalities is bracketed with F_Y(y)=...=f_Y(y), which is clearly not true. Have you missed a step in typing it out?

I suppose that we will never know ,

 

[vela]

Why t dt, not just dt? And your long line of equalities is bracketed with F_Y(y)=...=f_Y(y), which is clearly not true. Have you missed a step in typing it out?

They're not all equal signs; there's an arrow in there as well. The OP has $F_Y(y) = y/2\ \Rightarrow\ f_Y(y) = 1/2$, which is, in fact, okay.

 

[haruspex]

They're not all equal signs; there's an arrow in there as well. The OP has $F_Y(y) = y/2\ \Rightarrow\ f_Y(y) = 1/2$, which is, in fact, okay.

Ok. The ASCII character used for the arrow in the OP doesn't come out right on my iPad.