- #1
Barioth
- 49
- 0
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: