- #1
Addez123
- 199
- 21
- TL;DR Summary
- Find the .5 quantile (which is same as median for continuous functions?) given the distribution function:
$$f(x) = 2xe^{-x^2}$$
Given the distribution function
$$f(x) = 2xe^{-x^2}$$
The probability density function would then be
$$F(x) = -e^{-x^2}$$
To find the .5 quantile I set F(x) = .5
$$.5 = -e^{-x^2}$$
$$ln(-.5) = -x^2$$
And already here we have the issue, you can't take ln of a negative number.
$$f(x) = 2xe^{-x^2}$$
The probability density function would then be
$$F(x) = -e^{-x^2}$$
To find the .5 quantile I set F(x) = .5
$$.5 = -e^{-x^2}$$
$$ln(-.5) = -x^2$$
And already here we have the issue, you can't take ln of a negative number.