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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?

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  • $\begingroup$ I don't think so unless the two Weibull shape parameters are equal. $\endgroup$ Mar 11 '21 at 20:34
  • $\begingroup$ @Jarle I believe it can be expressed in terms of the incomplete Gamma function: the Weibull is just a power-transformed exponential ($\Gamma(1)$) distribution, after all. $\endgroup$
    – whuber
    Mar 11 '21 at 21:15
  • $\begingroup$ @whuber I agree but I don't see that it can be done if the two shape parameters are different. $\endgroup$ Mar 11 '21 at 23:23
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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 $$ P(X<x|X<Y)=1-\frac{\Gamma\left(\frac12,\left(\sqrt{a_2}x^b+\frac{a_1}{2\sqrt{a_2}}\right)^2\right)}{\Gamma\left(\frac12,\frac{a_1^2}{4a_2}\right)}. $$

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    $\begingroup$ I agree this is not easy, but some other special cases exist. For instance, I'm sure you can evaluate it when $k_y/k_x=2$ ;-). $\endgroup$
    – whuber
    Mar 11 '21 at 23:32

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