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Assume I have n independent random variables $\varepsilon_{i}$ with a Type I Extreme Value Distribution $F_{i}(c)=\exp (-\exp(-(c+x_{i})))$. I know that these are closed under maximization and therefore if $X \equiv max_{i}(\varepsilon_{i})$, the distribution of the maximum is just $$F_{x}(c)=\exp(-\exp(-(c-\log(\sum_{i}^{n}\exp(-x_{i}))))$$

Notice that $Prob(X\geq 0)=1-\exp(-\sum_{i}^{n}\exp(-x_{i}))$

Now, what I am trying to show as a substep for a probability I am trying to compute is that I can also compute the distribution of this $max$ using the Law of Total Probability, i.e

$$ Prob(X \geq 0)=\sum_{i=1}^{N}Prob(X>0|i=argmax_{j}\{ \varepsilon_{j} \}) \times Prob(i=argmax_{j}\{ \varepsilon_{j} \}) $$

$$ Prob(X \geq 0)=\sum_{i=1}^{N}Prob(\varepsilon_{i}>0) \frac{e^{-x_{i}}}{\sum_{j}e^{-x_{j}}} $$

$$ Prob(X \geq 0)= 1 - \frac{\sum_{i=1}^{N}e^{-(x_{i}+exp(-x_{i}))}}{\sum_{j}^{N}e^{-x_{j}}} $$

This is where I am stuck. Either I have a conceptual mistake and I can not use the Law of Total probability by some reason or I have an algebra mistake. If somebody could help me I would appreciate it a lot!

econ_ugrad
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1 Answers1

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You wrote:

$Prob(X \geq 0)=\sum_{i=1}^{N}Prob(X>0|i=argmax_{j}\{ \varepsilon_{j} \}) \times Prob(i=argmax_{j}\{ \varepsilon_{j} \})$

and

$Prob(X \geq 0)=\sum_{i=1}^{N}Prob(\varepsilon_{i}>0) \frac{e^{-x_{i}}}{\sum_{j}e^{-x_{j}}}$

Why do you think the following holds:

$Prob(i=argmax_{j}\{ \varepsilon_{j} \}) = \frac{e^{-x_{i}}}{\sum_{j}e^{-x_{j}}}$

Can you justify this step?

PTDS
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  • So the $Prob(i=argmax_{j} {\varepsilon_{j}})$ can be computed from the distribution. What is definitely wrong is that $Prob(X>0|i=argmax_{j} {\varepsilon_{j}})=Prob(\varepsilon_{i}>0)$ – econ_ugrad Aug 07 '19 at 13:59