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#### Barioth

##### Member

- Jan 17, 2013

- 52

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?

Thanks for reading

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?

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?

Thanks for reading

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: