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 Or maybe a way to approximate this probability?

• I don't think so unless the two Weibull shape parameters are equal. Mar 11 '21 at 20:34
• @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.
– whuber
Mar 11 '21 at 21:15
• @whuber I agree but I don't see that it can be done if the two shape parameters are different. Mar 11 '21 at 23:23

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 Similarly, $$P(X and so $$P(X
If instead $$b_1=b$$ and $$b_2=2b$$, \begin{align} P(X where $$\Gamma$$ is the (upper) incomplete Gamma function. A similar calculation for $$P(X leads to $$P(X
• I agree this is not easy, but some other special cases exist. For instance, I'm sure you can evaluate it when $k_y/k_x=2$ ;-).