A question on a contradiction by a random variable $X$ I am stuck on the following contradiction.
Suppose a positive random variable X satisfies the following.
$P(X>t)=(1-t)^2$
Then
$P(X^2>a)=(1-a^{1/2})^2$ holds.
Since $P(X^2=u)=P(X=u^{1/2})$, we get the following by differentiating and taking the minus sign.
$P(X=u^{1/2})=2-2u^{1/2}, P(X^2=u)=u^{-1/2}-1$
This is a clear contradiction, and I am not sure about


*

*Which specific part is wrong in this statement

*How to fix it


If anyone knows at least one of these questions, a kind answer would be appreciated. Thank you in advance!
+) The 'probability' actually means pdf, and the support of X should be [0, 1]
 A: You  (should) have: $$\mathsf P(X>t) = (1-t)^2\cdot\mathbf 1_{t\in[0;1]} + \mathbf 1_{t<0}$$
The probability density function (pdf) is then $$f(t)=\lvert\tfrac{\mathrm d~}{\mathrm d t}\mathsf P(X>t) \rvert = 2(1-t)\cdot\mathbf 1_{t\in[0;1]}$$
So then we can evaluate the pdf at $u^{1/2}$ as: $$f(u^{1/2}) = 2(1-u^{1/2})\cdot\mathbf 1_{u\in[0;1]}$$
Now if we wish:$$\begin{align}\mathsf P(X^2>u) &= \mathsf P(X>u^{1/2})\\ &= (1-u^{1/2})^2\cdot\mathbf 1_{u\in[0;1]}\end{align}$$
Then $$\begin{align}\lvert\tfrac{\mathrm d~}{\mathrm d u}\mathsf P(X^2>u) \rvert &= u^{-1/2}(1-u^{1/2})\cdot\mathbf 1_{u\in[0;1]} \\ &= (u^{-1/2}-1)\cdot\mathbf 1_{u\in[0;1]}\end{align}$$
Which does not equal the pdf evaluated at $u^{1/2}$, because of the chain rule for differentiation.   Indeed:
$$\left\lvert{\tfrac{\mathrm d~~~}{\mathrm d~u}\mathsf P(X^2>u) }\right\rvert~=~ \lvert\tfrac{\mathrm d~u^{1/2}}{\mathrm d~u~~~}\rvert~f(u^{1/2})~$$

PS: $\mathbf 1_{x\in A}$ is an indicator function; a piecewise function that is $1$ when the indicated condition is true, and $0$ otherwise. $$\mathbf 1_{x\in A}~=~\begin{cases}1 &:& x\in A\\ 0 &:& \text{otherwise}\end{cases}$$
A: For example, one wrong thing is that $P(X>t)=(1-t)^2>1$ when $t>2$ (probability greater than 1!).
Then, from your definition, it looks like $X$ is continuous random variable, then you talk about $P(X=u^{1/2})$ which is non-zero only for discrete random variable.
These two needs to be addressed before I would look into it any further.
A: First, I'll assume you are using $P(X^2 = u)$ as short hand for the pdf rather than actual probability.
One thing is that $P(X > t) = (1 - t)^2$ cannot be valid in general for a positive random variable. It must be the case that $0 \le t \le 1$.
$P(X < t) = 1 - P(X > t)$, so you should get
$P(X = u^{\frac{1}{2}}) = 1 - u^{\frac{1}{2}}.$ This is valid for $0 < u < 1$ which seems to be the domain of the random variable anyways.
