Find $E(text{min}(X_1,X_2,X_3))$ where each $X_i$ is exponential with parameter $i$
$begingroup$
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
edited Jan 15 at 23:03
Did
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
$endgroup$
add a comment |
$begingroup$
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
edited Jan 15 at 23:03
Did
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 15 at 23:03
Did
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
$endgroup$
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
edited Jan 15 at 23:03
Did
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
edited Jan 15 at 23:03
Did
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
edited Jan 15 at 23:03
Did
249k23226466
edited Jan 15 at 23:03
Did
249k23226466
edited Jan 15 at 23:03
Did
249k23226466
249k23226466
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
asked Jan 15 at 20:52
ParsevalParseval
3,0771719
3,0771719
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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votes
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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 $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.
answered Jan 15 at 21:08
ODFODF
1,486510
$endgroup$
add a comment |
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1 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 $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.
answered Jan 15 at 21:08
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
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 $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.
answered Jan 15 at 21:08
ODFODF
1,486510
$endgroup$
add a comment |
$begingroup$
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 $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.
answered Jan 15 at 21:08
ODFODF
1,486510
$endgroup$
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 $min(X_1,X_2,X_3) > t iff X_1, X_2, X_3 > t$.
answered Jan 15 at 21:08
ODFODF
1,486510
answered Jan 15 at 21:08
ODFODF
1,486510
answered Jan 15 at 21:08
ODFODF
1,486510
answered Jan 15 at 21:08
ODFODF
1,486510
1,486510
add a comment |
add a comment |
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$begingroup$
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