Find $E(text{min}(X_1,X_2,X_3))$ where each $X_i$ is exponential with parameter $i$












0












$begingroup$



I want to calculate $E(text{min}(X_1,X_2,X_3))$ where
$X_1simtext{exp}(1), X_2simtext{exp}(2)$ and
$X_3simtext{exp}(3).$




Denote $M=text{min}(X_1,X_2,X_3)$. By independence of the $X_i$ we have that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=1-mathbb{P}(M>t)=1-mathbb{P}(X_1>t)mathbb{P}(X_2>t)mathbb{P}(X_3>t)=1-e^{-6t},
end{align}



thus $f_M(t)=F_M'(t)=6e^{-6t}implies E(M)=1/6$.



Question:



Why isn't it the case that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=mathbb{P}(X_1leq t)mathbb{P}(X_2leq t)mathbb{P}(X_3leq t)=(1-e^{-x})(1-e^{-2x})(1-e^{-3x})\
&=... ?
end{align}



Why do I have to use the complement rule?






probability-theory exponential-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why the wrong title?
    $endgroup$
    – Did
    Jan 15 at 23:03










  • $begingroup$
    @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
    $endgroup$
    – Parseval
    Jan 15 at 23:05










  • $begingroup$
    What was your title and what is it now that I have corrected it? Tired or not...
    $endgroup$
    – Did
    Jan 15 at 23:07










  • $begingroup$
    @Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
    $endgroup$
    – Parseval
    Jan 15 at 23:31
















0












$begingroup$



I want to calculate $E(text{min}(X_1,X_2,X_3))$ where
$X_1simtext{exp}(1), X_2simtext{exp}(2)$ and
$X_3simtext{exp}(3).$




Denote $M=text{min}(X_1,X_2,X_3)$. By independence of the $X_i$ we have that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=1-mathbb{P}(M>t)=1-mathbb{P}(X_1>t)mathbb{P}(X_2>t)mathbb{P}(X_3>t)=1-e^{-6t},
end{align}



thus $f_M(t)=F_M'(t)=6e^{-6t}implies E(M)=1/6$.



Question:



Why isn't it the case that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=mathbb{P}(X_1leq t)mathbb{P}(X_2leq t)mathbb{P}(X_3leq t)=(1-e^{-x})(1-e^{-2x})(1-e^{-3x})\
&=... ?
end{align}



Why do I have to use the complement rule?






probability-theory exponential-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why the wrong title?
    $endgroup$
    – Did
    Jan 15 at 23:03










  • $begingroup$
    @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
    $endgroup$
    – Parseval
    Jan 15 at 23:05










  • $begingroup$
    What was your title and what is it now that I have corrected it? Tired or not...
    $endgroup$
    – Did
    Jan 15 at 23:07










  • $begingroup$
    @Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
    $endgroup$
    – Parseval
    Jan 15 at 23:31














0












0








0





$begingroup$



I want to calculate $E(text{min}(X_1,X_2,X_3))$ where
$X_1simtext{exp}(1), X_2simtext{exp}(2)$ and
$X_3simtext{exp}(3).$




Denote $M=text{min}(X_1,X_2,X_3)$. By independence of the $X_i$ we have that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=1-mathbb{P}(M>t)=1-mathbb{P}(X_1>t)mathbb{P}(X_2>t)mathbb{P}(X_3>t)=1-e^{-6t},
end{align}



thus $f_M(t)=F_M'(t)=6e^{-6t}implies E(M)=1/6$.



Question:



Why isn't it the case that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=mathbb{P}(X_1leq t)mathbb{P}(X_2leq t)mathbb{P}(X_3leq t)=(1-e^{-x})(1-e^{-2x})(1-e^{-3x})\
&=... ?
end{align}



Why do I have to use the complement rule?






probability-theory exponential-distribution







share|cite|improve this question











$endgroup$





I want to calculate $E(text{min}(X_1,X_2,X_3))$ where
$X_1simtext{exp}(1), X_2simtext{exp}(2)$ and
$X_3simtext{exp}(3).$




Denote $M=text{min}(X_1,X_2,X_3)$. By independence of the $X_i$ we have that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=1-mathbb{P}(M>t)=1-mathbb{P}(X_1>t)mathbb{P}(X_2>t)mathbb{P}(X_3>t)=1-e^{-6t},
end{align}



thus $f_M(t)=F_M'(t)=6e^{-6t}implies E(M)=1/6$.



Question:



Why isn't it the case that



begin{align}
F_M(t)&=mathbb{P}(Mleq t)=mathbb{P}(X_1leq t)mathbb{P}(X_2leq t)mathbb{P}(X_3leq t)=(1-e^{-x})(1-e^{-2x})(1-e^{-3x})\
&=... ?
end{align}



Why do I have to use the complement rule?






probability-theory exponential-distribution




probability-theory exponential-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 23:03









Did

247k23223460




247k23223460










asked Jan 15 at 20:52









ParsevalParseval

2,8821718




2,8821718












  • $begingroup$
    Why the wrong title?
    $endgroup$
    – Did
    Jan 15 at 23:03










  • $begingroup$
    @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
    $endgroup$
    – Parseval
    Jan 15 at 23:05










  • $begingroup$
    What was your title and what is it now that I have corrected it? Tired or not...
    $endgroup$
    – Did
    Jan 15 at 23:07










  • $begingroup$
    @Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
    $endgroup$
    – Parseval
    Jan 15 at 23:31


















  • $begingroup$
    Why the wrong title?
    $endgroup$
    – Did
    Jan 15 at 23:03










  • $begingroup$
    @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
    $endgroup$
    – Parseval
    Jan 15 at 23:05










  • $begingroup$
    What was your title and what is it now that I have corrected it? Tired or not...
    $endgroup$
    – Did
    Jan 15 at 23:07










  • $begingroup$
    @Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
    $endgroup$
    – Parseval
    Jan 15 at 23:31
















$begingroup$
Why the wrong title?
$endgroup$
– Did
Jan 15 at 23:03




$begingroup$
Why the wrong title?
$endgroup$
– Did
Jan 15 at 23:03












$begingroup$
@Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
$endgroup$
– Parseval
Jan 15 at 23:05




$begingroup$
@Did - I'm sorry I'm a bit tired now. I don't see what you're refering to?
$endgroup$
– Parseval
Jan 15 at 23:05












$begingroup$
What was your title and what is it now that I have corrected it? Tired or not...
$endgroup$
– Did
Jan 15 at 23:07




$begingroup$
What was your title and what is it now that I have corrected it? Tired or not...
$endgroup$
– Did
Jan 15 at 23:07












$begingroup$
@Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
$endgroup$
– Parseval
Jan 15 at 23:31




$begingroup$
@Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them.
$endgroup$
– Parseval
Jan 15 at 23:31










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Because $Bbb{P}(Mleq t) neq Bbb{P}(X_1 leq t)Bbb{P}(X_2 leq t)Bbb{P}(X_3 leq t)$. Indeed - if $X_1 leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M leq t$.



However, it is not too hard to see that $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    Because $Bbb{P}(Mleq t) neq Bbb{P}(X_1 leq t)Bbb{P}(X_2 leq t)Bbb{P}(X_3 leq t)$. Indeed - if $X_1 leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M leq t$.



    However, it is not too hard to see that $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Because $Bbb{P}(Mleq t) neq Bbb{P}(X_1 leq t)Bbb{P}(X_2 leq t)Bbb{P}(X_3 leq t)$. Indeed - if $X_1 leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M leq t$.



      However, it is not too hard to see that $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Because $Bbb{P}(Mleq t) neq Bbb{P}(X_1 leq t)Bbb{P}(X_2 leq t)Bbb{P}(X_3 leq t)$. Indeed - if $X_1 leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M leq t$.



        However, it is not too hard to see that $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.






        share|cite|improve this answer









        $endgroup$



        Because $Bbb{P}(Mleq t) neq Bbb{P}(X_1 leq t)Bbb{P}(X_2 leq t)Bbb{P}(X_3 leq t)$. Indeed - if $X_1 leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M leq t$.



        However, it is not too hard to see that $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 21:08









        ODFODF

        1,486510




        1,486510






























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