# Uniforme [0,1] squared

#### Barioth

##### Member
Hi everyone!

Here is my question:

Let's say U a continuous random variable, U is a uniform [0,1]

We're looking for $$\displaystyle U^2$$ Density.

I go with

$$\displaystyle P(U^2<a)=P(U<a^{1/2})$$

Altough my teacher say I must go with

$$\displaystyle P(U^2<a)=P(-a^{1/2}<U<a^{1/2})$$

If we've U in [0,1] I don't see why we would want to look at value that are under 0?

Edit: Thinking about it, it is actualy the same since we can break it as

$$\displaystyle P(U^2<a)=P(-a^{1/2}<U<a^{1/2}) =P(-a^{1/2}<U<0)+P(0<=U<a^{1/2})$$
$$\displaystyle = 0 + P(0<U<a^{1/2})= P(U<a^{1/2})$$

Am I right?

Last edited:

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Yes. You are right.

#### Barioth

##### Member
Thanks, it was in my exam last week, the teacher gave me my point back, a great teacher