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I have been working on this for a while now, but unfortunately I am not getting further with this, so I was hoping for some help maybe.

Consider two independent continuous random variables $X_1$ and $X_2$ on the probability space $(\Omega, \mathcal{F},\mathcal{P})$ with positive support and density functions $f_1(x,\theta_1)$ and $f_2(x,\theta_2)$ respectively. Also consider a discrete indicator random variable $C=1$ if $X_1 < X_2$ and $C=0$ otherwise.

I would like to compute the conditional probabilities $P(C=c|Z=z)$, $c \in \{0,1\}$, where $Z:= \min \{X_1,X_2\}$.

I tried the following: After some computation, we get

  1. $f_{Z,C}(z,c) = [f_1(z)\overline{F}_2(z)]^c * [f_2(z)\overline{F}_1(z)]^{1-c} $
  2. $f_Z(z)=f_1(z)\overline{F}_2(z) + f_2(z)\overline{F}_1(z),$

where $\overline{F}$ denotes the survival function. Hence:

$P(C=c|Z=z)= \frac{f_{Z,C}(z,c)}{f_Z(y)} = \frac{[f_1(z)\overline{F}_2(z)]^c * [f_2(z)\overline{F}_1(z)]^{1-c}}{f_1(z)\overline{F}_2(z) + f_2(z)\overline{F}_1(z)}$

However, I don't believe that this is correct. Take $X_1\sim Exp(\lambda_1)$ and $X_2\sim Exp(\lambda_2)$, with density functions $f_1(x)=\lambda_1 e^{-\lambda_1 x}$ and $f_2$ correspondingly for instance. Then:

$P(C=c|Z=z)= \frac{(\lambda_1 e^{-(\lambda_1+\lambda_2)z})^{c} * (\lambda_2 e^{-(\lambda_1+\lambda_2)z})^{1-c}}{\lambda_1 e^{-(\lambda_1+\lambda_2)z} + \lambda_2 e^{-(\lambda_1+\lambda_2)z}}=\frac{\lambda_1^c * \lambda_2^{1-c}}{\lambda_1 +\lambda_2}$,

which is indepedent of $z$. This seems wrong to me as one would expect the probability of the minimum to be a realisation of $X_1$ or $X_2$ to depend on where that realisation occurred. At least this is my thinking. Let me know if you can spot where I went wrong please. Greatly appreciate any help! Thanks!

RPP
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1 Answers1

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No. Your calculations are correct.

You just chose to test with a situation where the event that $X_1$ is the minimum really is independent of the value of the minimum, due to the geometry of their joint distribution.

Try choosing distributions with less symmetry, such as with triangle distributed random variables where $f_1(x)=2x\mathbf 1_{0\leqslant x\leqslant 1}$ and $f_2(x)=2(1-x)\mathbf 1_{0\leqslant x\leqslant 1}$.

Graham Kemp
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    Thank you Graham. I appreciate your insights. The results of the propsed distributions with less symmetry are in line with my intuition a lot better. For example P(C=1|Z=z) < 1/3 for all z in (0,1), which is reasonable as f_2 represents the 'more sensitive' lifetime. I am still troubled by the original example though despite the calculations being right. I guess it does come down to the particular geometry of the pdf's but my intuition just can't accept that (yet) (-; Again, thanks so much for your help. Really appreciate it! – RPP Feb 05 '22 at 13:25