Compute the probability $P(X ≤ Y|min{X,Y}=x)$












1












$begingroup$


Let $X$ and $Y$ be two independent exponential random variables with parameters $lambda$ and $mu$, respectively. Compute the probability $P(X ≤ Y|min{X,Y}=x)$.



Here is what I did:

If $Y=min{X,Y}$, then $Ylt X$, $P(X ≤ Y|min{X,Y}=x)=0$.

If $X=min{X,Y}$, then $Xlt Y$,
$$
begin{aligned}
P(X ≤ Y|min{X,Y}=x)&=P(X ≤ Y|X=x)\
&=frac{P(Ygt x)}{f_X(x)}\
&=frac{1-P(Ylt x)}{f_X(x)}\
&={1-int_{0}^xmu e^{-mu y}dyover lambda e^{-lambda x}}
end{aligned}
$$



Am I right?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
    $endgroup$
    – Xiaomi
    Oct 1 '18 at 1:45










  • $begingroup$
    @Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:52










  • $begingroup$
    @YibeiHe, still no. $f_X(x) ne P(X=x)$
    $endgroup$
    – Zamarion
    Oct 1 '18 at 1:54


















1












$begingroup$


Let $X$ and $Y$ be two independent exponential random variables with parameters $lambda$ and $mu$, respectively. Compute the probability $P(X ≤ Y|min{X,Y}=x)$.



Here is what I did:

If $Y=min{X,Y}$, then $Ylt X$, $P(X ≤ Y|min{X,Y}=x)=0$.

If $X=min{X,Y}$, then $Xlt Y$,
$$
begin{aligned}
P(X ≤ Y|min{X,Y}=x)&=P(X ≤ Y|X=x)\
&=frac{P(Ygt x)}{f_X(x)}\
&=frac{1-P(Ylt x)}{f_X(x)}\
&={1-int_{0}^xmu e^{-mu y}dyover lambda e^{-lambda x}}
end{aligned}
$$



Am I right?






probability







share|cite|improve this question











$endgroup$












  • $begingroup$
    No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
    $endgroup$
    – Xiaomi
    Oct 1 '18 at 1:45










  • $begingroup$
    @Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:52










  • $begingroup$
    @YibeiHe, still no. $f_X(x) ne P(X=x)$
    $endgroup$
    – Zamarion
    Oct 1 '18 at 1:54
















1












1








1


3



$begingroup$


Let $X$ and $Y$ be two independent exponential random variables with parameters $lambda$ and $mu$, respectively. Compute the probability $P(X ≤ Y|min{X,Y}=x)$.



Here is what I did:

If $Y=min{X,Y}$, then $Ylt X$, $P(X ≤ Y|min{X,Y}=x)=0$.

If $X=min{X,Y}$, then $Xlt Y$,
$$
begin{aligned}
P(X ≤ Y|min{X,Y}=x)&=P(X ≤ Y|X=x)\
&=frac{P(Ygt x)}{f_X(x)}\
&=frac{1-P(Ylt x)}{f_X(x)}\
&={1-int_{0}^xmu e^{-mu y}dyover lambda e^{-lambda x}}
end{aligned}
$$



Am I right?






probability







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ be two independent exponential random variables with parameters $lambda$ and $mu$, respectively. Compute the probability $P(X ≤ Y|min{X,Y}=x)$.



Here is what I did:

If $Y=min{X,Y}$, then $Ylt X$, $P(X ≤ Y|min{X,Y}=x)=0$.

If $X=min{X,Y}$, then $Xlt Y$,
$$
begin{aligned}
P(X ≤ Y|min{X,Y}=x)&=P(X ≤ Y|X=x)\
&=frac{P(Ygt x)}{f_X(x)}\
&=frac{1-P(Ylt x)}{f_X(x)}\
&={1-int_{0}^xmu e^{-mu y}dyover lambda e^{-lambda x}}
end{aligned}
$$



Am I right?






probability




probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 1 '18 at 1:51







Yibei He

















asked Oct 1 '18 at 1:29









Yibei HeYibei He

3139




3139












  • $begingroup$
    No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
    $endgroup$
    – Xiaomi
    Oct 1 '18 at 1:45










  • $begingroup$
    @Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:52










  • $begingroup$
    @YibeiHe, still no. $f_X(x) ne P(X=x)$
    $endgroup$
    – Zamarion
    Oct 1 '18 at 1:54




















  • $begingroup$
    No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
    $endgroup$
    – Xiaomi
    Oct 1 '18 at 1:45










  • $begingroup$
    @Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:52










  • $begingroup$
    @YibeiHe, still no. $f_X(x) ne P(X=x)$
    $endgroup$
    – Zamarion
    Oct 1 '18 at 1:54


















$begingroup$
No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
$endgroup$
– Xiaomi
Oct 1 '18 at 1:45




$begingroup$
No. You should answer in a single statement, not a case by case basis. Secondly, your proof is wrong because $P(X = x)$ is $0$ since $X$ is a continuous random variable, so you're dividing by zero
$endgroup$
– Xiaomi
Oct 1 '18 at 1:45












$begingroup$
@Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
$endgroup$
– Yibei He
Oct 1 '18 at 1:52




$begingroup$
@Xiaomi I improved my answer. Is that correct now? And how to answer in a single statement?
$endgroup$
– Yibei He
Oct 1 '18 at 1:52












$begingroup$
@YibeiHe, still no. $f_X(x) ne P(X=x)$
$endgroup$
– Zamarion
Oct 1 '18 at 1:54






$begingroup$
@YibeiHe, still no. $f_X(x) ne P(X=x)$
$endgroup$
– Zamarion
Oct 1 '18 at 1:54












2 Answers
2






active

oldest

votes


















1












$begingroup$

$$P(X le Y mid min{X, Y} = x) = frac{P(X le Y, min{X,Y} in [x, x+dx])}{P(min{X,Y} in [x, x+dx])}.$$



The numerator is equal to
$$P(X le Y, X in [x, x + dx]) = P(X in [x, x+dx]) P(Y ge x) = (lambda e^{-lambda x} , dx)(e^{-mu x}).$$



The denominator is equal to
$$P(X le Y, X in [x, x + dx]) + P(X ge Y, Y in [x, x + dx])
= (lambda e^{-lambda x} , dx)(e^{-mu x}) + (mu e^{-mu x} , dx) (e^{-lambda x})$$
by similar computations.



Thus the answer is $frac{lambda}{lambda + mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]




Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-lambda x} - e^{-lambda (x + dx)} = e^{-lambda x}(1 - e^{-lambda , dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-lambda dx} = 1 - lambda , dx$ to obtain $lambda e^{-lambda x} , dx$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 2:28










  • $begingroup$
    @YibeiHe See my edit.
    $endgroup$
    – angryavian
    Oct 1 '18 at 5:56



















1












$begingroup$

The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.



You can compute this probability that way,

begin{align}
P(X<Y|X=x) &= P(x<Y) \
&= 1 - int_0^xmu e^{-mu y}dy \
&= 1 - (1-e^{-mu x}) \
&= e^{mu x}
end{align}



Note that this is not the final answer and you should be careful on how you have split your two cases.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Shall I divide it by the pdf of X since the condition is X=x.
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:54











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

$$P(X le Y mid min{X, Y} = x) = frac{P(X le Y, min{X,Y} in [x, x+dx])}{P(min{X,Y} in [x, x+dx])}.$$



The numerator is equal to
$$P(X le Y, X in [x, x + dx]) = P(X in [x, x+dx]) P(Y ge x) = (lambda e^{-lambda x} , dx)(e^{-mu x}).$$



The denominator is equal to
$$P(X le Y, X in [x, x + dx]) + P(X ge Y, Y in [x, x + dx])
= (lambda e^{-lambda x} , dx)(e^{-mu x}) + (mu e^{-mu x} , dx) (e^{-lambda x})$$
by similar computations.



Thus the answer is $frac{lambda}{lambda + mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]




Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-lambda x} - e^{-lambda (x + dx)} = e^{-lambda x}(1 - e^{-lambda , dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-lambda dx} = 1 - lambda , dx$ to obtain $lambda e^{-lambda x} , dx$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 2:28










  • $begingroup$
    @YibeiHe See my edit.
    $endgroup$
    – angryavian
    Oct 1 '18 at 5:56
















1












$begingroup$

$$P(X le Y mid min{X, Y} = x) = frac{P(X le Y, min{X,Y} in [x, x+dx])}{P(min{X,Y} in [x, x+dx])}.$$



The numerator is equal to
$$P(X le Y, X in [x, x + dx]) = P(X in [x, x+dx]) P(Y ge x) = (lambda e^{-lambda x} , dx)(e^{-mu x}).$$



The denominator is equal to
$$P(X le Y, X in [x, x + dx]) + P(X ge Y, Y in [x, x + dx])
= (lambda e^{-lambda x} , dx)(e^{-mu x}) + (mu e^{-mu x} , dx) (e^{-lambda x})$$
by similar computations.



Thus the answer is $frac{lambda}{lambda + mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]




Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-lambda x} - e^{-lambda (x + dx)} = e^{-lambda x}(1 - e^{-lambda , dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-lambda dx} = 1 - lambda , dx$ to obtain $lambda e^{-lambda x} , dx$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 2:28










  • $begingroup$
    @YibeiHe See my edit.
    $endgroup$
    – angryavian
    Oct 1 '18 at 5:56














1












1








1





$begingroup$

$$P(X le Y mid min{X, Y} = x) = frac{P(X le Y, min{X,Y} in [x, x+dx])}{P(min{X,Y} in [x, x+dx])}.$$



The numerator is equal to
$$P(X le Y, X in [x, x + dx]) = P(X in [x, x+dx]) P(Y ge x) = (lambda e^{-lambda x} , dx)(e^{-mu x}).$$



The denominator is equal to
$$P(X le Y, X in [x, x + dx]) + P(X ge Y, Y in [x, x + dx])
= (lambda e^{-lambda x} , dx)(e^{-mu x}) + (mu e^{-mu x} , dx) (e^{-lambda x})$$
by similar computations.



Thus the answer is $frac{lambda}{lambda + mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]




Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-lambda x} - e^{-lambda (x + dx)} = e^{-lambda x}(1 - e^{-lambda , dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-lambda dx} = 1 - lambda , dx$ to obtain $lambda e^{-lambda x} , dx$.






share|cite|improve this answer











$endgroup$



$$P(X le Y mid min{X, Y} = x) = frac{P(X le Y, min{X,Y} in [x, x+dx])}{P(min{X,Y} in [x, x+dx])}.$$



The numerator is equal to
$$P(X le Y, X in [x, x + dx]) = P(X in [x, x+dx]) P(Y ge x) = (lambda e^{-lambda x} , dx)(e^{-mu x}).$$



The denominator is equal to
$$P(X le Y, X in [x, x + dx]) + P(X ge Y, Y in [x, x + dx])
= (lambda e^{-lambda x} , dx)(e^{-mu x}) + (mu e^{-mu x} , dx) (e^{-lambda x})$$
by similar computations.



Thus the answer is $frac{lambda}{lambda + mu}$. Note that it is interesting that the answer does not depend on $x$. [You could also quickly answer this question using properties of Poisson processes, but perhaps this is off-topic.]




Response to comment: It actually does work. $$F(x + dx) - F(x) = e^{-lambda x} - e^{-lambda (x + dx)} = e^{-lambda x}(1 - e^{-lambda , dx}).$$
Then using $e^z = 1 + z + O(z^2)$, we can plug in $e^{-lambda dx} = 1 - lambda , dx$ to obtain $lambda e^{-lambda x} , dx$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 1 '18 at 5:56

























answered Oct 1 '18 at 1:55









angryavianangryavian

42.1k23381




42.1k23381












  • $begingroup$
    Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 2:28










  • $begingroup$
    @YibeiHe See my edit.
    $endgroup$
    – angryavian
    Oct 1 '18 at 5:56


















  • $begingroup$
    Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
    $endgroup$
    – Yibei He
    Oct 1 '18 at 2:28










  • $begingroup$
    @YibeiHe See my edit.
    $endgroup$
    – angryavian
    Oct 1 '18 at 5:56
















$begingroup$
Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
$endgroup$
– Yibei He
Oct 1 '18 at 2:28




$begingroup$
Thanks a lot! One question: if I do the part P(x=[x,x+dx]) into F(x+dx)-F(x), it doesn’t work. Why?
$endgroup$
– Yibei He
Oct 1 '18 at 2:28












$begingroup$
@YibeiHe See my edit.
$endgroup$
– angryavian
Oct 1 '18 at 5:56




$begingroup$
@YibeiHe See my edit.
$endgroup$
– angryavian
Oct 1 '18 at 5:56











1












$begingroup$

The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.



You can compute this probability that way,

begin{align}
P(X<Y|X=x) &= P(x<Y) \
&= 1 - int_0^xmu e^{-mu y}dy \
&= 1 - (1-e^{-mu x}) \
&= e^{mu x}
end{align}



Note that this is not the final answer and you should be careful on how you have split your two cases.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Shall I divide it by the pdf of X since the condition is X=x.
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:54
















1












$begingroup$

The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.



You can compute this probability that way,

begin{align}
P(X<Y|X=x) &= P(x<Y) \
&= 1 - int_0^xmu e^{-mu y}dy \
&= 1 - (1-e^{-mu x}) \
&= e^{mu x}
end{align}



Note that this is not the final answer and you should be careful on how you have split your two cases.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Shall I divide it by the pdf of X since the condition is X=x.
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:54














1












1








1





$begingroup$

The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.



You can compute this probability that way,

begin{align}
P(X<Y|X=x) &= P(x<Y) \
&= 1 - int_0^xmu e^{-mu y}dy \
&= 1 - (1-e^{-mu x}) \
&= e^{mu x}
end{align}



Note that this is not the final answer and you should be careful on how you have split your two cases.






share|cite|improve this answer











$endgroup$



The second part isn't quite right, since $P(X=x)=0$ for any continuous random variable $X$.



You can compute this probability that way,

begin{align}
P(X<Y|X=x) &= P(x<Y) \
&= 1 - int_0^xmu e^{-mu y}dy \
&= 1 - (1-e^{-mu x}) \
&= e^{mu x}
end{align}



Note that this is not the final answer and you should be careful on how you have split your two cases.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 1 '18 at 2:28

























answered Oct 1 '18 at 1:47









ZamarionZamarion

33719




33719












  • $begingroup$
    Shall I divide it by the pdf of X since the condition is X=x.
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:54


















  • $begingroup$
    Shall I divide it by the pdf of X since the condition is X=x.
    $endgroup$
    – Yibei He
    Oct 1 '18 at 1:54
















$begingroup$
Shall I divide it by the pdf of X since the condition is X=x.
$endgroup$
– Yibei He
Oct 1 '18 at 1:54




$begingroup$
Shall I divide it by the pdf of X since the condition is X=x.
$endgroup$
– Yibei He
Oct 1 '18 at 1:54


















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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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