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If I know the CDF of $X$ where $X=\min(X_1,X_2)$, is it possible to find the CDFs of $X_1$ and $X_2$? I know that $X_1$ and $X_2$ are i.i.d. such that $$F_{X}(x)=2F_{X_i}(x)-F_{X_i}(x)^2$$ where $i=$ 1 or 2.

It may be difficult to solve this equation. Is there any systematic way for this kind of problem?

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If $X_1$ and $X_2$ are iid, then your expression for $X=\operatorname{min}(X_1,X_2)$ always holds. This is because of the identical distribution and independence: $F_{X} (x)=\mathbb{P}(X\leq x)= 1- \mathbb{P}(X>x)=1-\mathbb{P}(X_1> x, X_2> x)= \\ =1-\mathbb{P}(X_1> x) \mathbb{P}(X_2> x) = 1- [1-F_{X_i}(x)]^2 = \\ = 2F_{X_i}(x) - F_{X_i}(x)^2$.

Now, if you are given an expression of $F_{X} (x)$ as a function of $x$, you could solve the equation above as a quadratic equation for $F_{X_i}(x)$ and find the corresponding cdf and distribution, again as a function of $x$. Hope this helps.

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Define $G_X(x) := 1- F_X(x)$. Likewise, $G_{X_i}(x) := 1-F_{X_i}(x)$ for $i\in\{1,2\}$. Given that $X_1$ and $X_2$ are iid and $X = \min(X_1, X_2)$,

$G_X(x) = \Pr(X > x) = \Pr(\min(X_1, X_2) > x) = \Pr(X_1 > x, X_2 > x) = \Pr(X_1 > x)\Pr(X_2 > x) = (G_{X_1}(x))(G_{X_2}(x)) =(G_{X_1}(x))^2$

Therefore, $G_{X_1}(x) = \sqrt{G_X(x)}$.

Consequently, $F_{X_1}(x) = 1-\sqrt{1-F_X(x)}$.

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