I like Serena said:Hi number0!
The integral in your solution runs from -√x to +√x, but that is only true if they are within the bounds of your uniform distribution.
If they are outside, you have values of x for which your integral bounds need to be modified, which in turn will result in other values for your density function.
If you do this, you will find your boundaries for U^2 implicitly.
number0 said:
I like Serena said:
"Uniform on [-1, 1]" means that U takes on values between -1 and 1 with equal probability. In other words, the probability of U being any number between -1 and 1 is the same.
The density function of U^2 can help us understand the distribution of U^2, which is useful in many statistical and mathematical applications. It can also help us calculate probabilities and make predictions about the behavior of U^2.
The density function of U^2 is related to the density function of U through a transformation. In this case, we are taking the square of U, so the density function of U^2 will be related to the density function of U through a squared term.
The formula for the density function of U^2 is f(x) = 1/(2*sqrt(x)), where x is between 0 and 1. This formula can be derived using the transformation method and the fact that U is uniform on [-1,1].
No, the density function of U^2 is specifically for when U is uniform on [-1, 1]. It cannot be used for other distributions without making additional adjustments or transformations.
