<s>[INST] Probability of Weibull RV conditional on another Weibull RV Let $X, Y$ be independent Weibull distributed random variables and $x>0$ a constant. Is there a closed form solution to calculating the probability $$P(X<x|X<Y)?$$
Or maybe a way to approximate this probability?
 [/INST] Letting $a_1,b_1$ and $a_2,b_2$ denote the parameters of $X$ and $Y$, and assuming that $b_1=b_2=b$,
\begin{align}
P(X<x \cap X<Y)
  &=\int_0^x \int_x^\infty f_X(x)f_Y(y) dy\,dx
\\&=\int_0^x f_X(x)(1-F_Y(x))dx
\\&=\int_0^x a_1 b x^{b-1}e^{-a_1 x^b - a_2 x^b} dx
\\&=\frac{a_1}{a_1+a_2}\int_0^{(a_1+a_2)x^b}e^{-u}du
\\&=\frac{a_1}{a_1+a_2}(1-e^{-(a_1+a_2)x^b}).
\end{align}
Similarly,
$$
P(X<Y)=\int_0^\infty \int_x^\infty f_X(x)f_Y(y) dy\,dx=\frac{a_1}{a_1+a_2}
$$
and so
$$
P(X<x|X<Y)=1-e^{-(a_1+a_2)x^b}.
$$
If instead $b_1=b$ and $b_2=2b$,
\begin{align}
P(X<x \cap X<Y)
  &=\int_0^x a_1 b x^{b-1}e^{-a_1 x^b - a_2 x^{2b}} dx
\\&=\frac{a_1}{\sqrt{a_2}}\int_0^{\sqrt{a_2}x^b}e^{-(u^2+\frac{a_1}{\sqrt{a_2}}u)}du
\\&=\frac{a_1}{\sqrt{a_2}}e^{\frac{a_1^2}{4a_2}}\int_0^{\sqrt{a_2}x^b}e^{-(u+\frac{a_1}{2\sqrt{a_2}})^2}du
\\&=\frac{a_1}{\sqrt{a_2}}e^{\frac{a_1^2}{4a_2}}\int_{\frac{a_1^2}{4a_2}}^{(\sqrt{a_2}x^b+\frac{a_1}{2\sqrt{a_2}})^2}v^{-\frac12}e^{-v}dv
\\&=\frac{a_1}{\sqrt{a_2}}e^{\frac{a_1^2}{4a_2}}\left(\Gamma\left(\frac12,\frac{a_1^2}{4a_2}\right)-\Gamma\left(\frac12,\left(\sqrt{a_2}x^b+\frac{a_1}{2\sqrt{a_2}}\right)^2\right)\right),
\end{align}
where $\Gamma$ is the (upper) incomplete Gamma function.  A similar calculation for $P(X<Y)$ leads to
$$
P(X<x|X<Y)=1-\frac{\Gamma\left(\frac12,\left(\sqrt{a_2}x^b+\frac{a_1}{2\sqrt{a_2}}\right)^2\right)}{\Gamma\left(\frac12,\frac{a_1^2}{4a_2}\right)}.
$$
</s>