joint PDF of max and min of $n$ iid standard uniform random variables
$begingroup$
Let $U_1, ... U_n$ be iid standard uniform variables.
Let $X = max(U_i)$ and $Y = min(U_i)$.
The goal is to compute the joint PDF of $X, Y$!
I have already computed the PDFs of $X$ and $Y$ separately.
But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.
I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X leq x, Y leq y) = P(U_1, ...U_n leq x, text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 geq y, ... U_n geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.
probability probability-theory probability-distributions uniform-distribution
asked Dec 13 '18 at 2:15
0k330k33
12210
$endgroup$
add a comment |
$begingroup$
Let $U_1, ... U_n$ be iid standard uniform variables.
Let $X = max(U_i)$ and $Y = min(U_i)$.
The goal is to compute the joint PDF of $X, Y$!
I have already computed the PDFs of $X$ and $Y$ separately.
But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.
I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X leq x, Y leq y) = P(U_1, ...U_n leq x, text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 geq y, ... U_n geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.
probability probability-theory probability-distributions uniform-distribution
asked Dec 13 '18 at 2:15
0k330k33
12210
$endgroup$
add a comment |
$begingroup$
Let $U_1, ... U_n$ be iid standard uniform variables.
Let $X = max(U_i)$ and $Y = min(U_i)$.
The goal is to compute the joint PDF of $X, Y$!
I have already computed the PDFs of $X$ and $Y$ separately.
But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.
I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X leq x, Y leq y) = P(U_1, ...U_n leq x, text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 geq y, ... U_n geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.
probability probability-theory probability-distributions uniform-distribution
asked Dec 13 '18 at 2:15
0k330k33
12210
$endgroup$
Let $U_1, ... U_n$ be iid standard uniform variables.
Let $X = max(U_i)$ and $Y = min(U_i)$.
The goal is to compute the joint PDF of $X, Y$!
I have already computed the PDFs of $X$ and $Y$ separately.
But I am not sure if that is so useful because $X, Y$ are not independent (by definition $X geq Y$) so it is not correct to just multiply the marginal PDFs of $X$ and $Y$.
I thought about starting from the CDF and taking advantage of the fact that all of the $U_i$'s are independent. So something like: $P(X leq x, Y leq y) = P(U_1, ...U_n leq x, text{at least one $U_i$ is less than }y)$ and you can already see the problem here -- I don't know how to express the relation for $Y$ in terms of all the $U_i$'s like that. when computing $Y$'s CDF, I did $1 - P(U_1 geq y, ... U_n geq y)$ and was able to take advantage of the independence of all the $U_i$'s there. But I'm not sure how to translate that to the joint PDF of $X, Y$.
probability probability-theory probability-distributions uniform-distribution
probability probability-theory probability-distributions uniform-distribution
asked Dec 13 '18 at 2:15
0k330k33
12210
asked Dec 13 '18 at 2:15
0k330k33
12210
asked Dec 13 '18 at 2:15
0k330k33
12210
asked Dec 13 '18 at 2:15
0k330k33
12210
asked Dec 13 '18 at 2:15
0k330k33
12210
12210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, on $A={0le y<xle 1}$,
begin{align}
mathsf{P}(Xle x,Yle y)&=mathsf{P}(Xle x)-mathsf{P}(Y>y,Xle x) \
&=mathsf{P}left(bigcap_{1le ile n}(U_ile x)right)-mathsf{P}left(bigcap_{1le ile n}(y<U_ile x)right) \
&=x^n-(x-y)^n.
end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
$endgroup$
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
First, on $A={0le y<xle 1}$,
begin{align}
mathsf{P}(Xle x,Yle y)&=mathsf{P}(Xle x)-mathsf{P}(Y>y,Xle x) \
&=mathsf{P}left(bigcap_{1le ile n}(U_ile x)right)-mathsf{P}left(bigcap_{1le ile n}(y<U_ile x)right) \
&=x^n-(x-y)^n.
end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
$endgroup$
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
add a comment |
$begingroup$
First, on $A={0le y<xle 1}$,
begin{align}
mathsf{P}(Xle x,Yle y)&=mathsf{P}(Xle x)-mathsf{P}(Y>y,Xle x) \
&=mathsf{P}left(bigcap_{1le ile n}(U_ile x)right)-mathsf{P}left(bigcap_{1le ile n}(y<U_ile x)right) \
&=x^n-(x-y)^n.
end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
$endgroup$
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
add a comment |
$begingroup$
First, on $A={0le y<xle 1}$,
begin{align}
mathsf{P}(Xle x,Yle y)&=mathsf{P}(Xle x)-mathsf{P}(Y>y,Xle x) \
&=mathsf{P}left(bigcap_{1le ile n}(U_ile x)right)-mathsf{P}left(bigcap_{1le ile n}(y<U_ile x)right) \
&=x^n-(x-y)^n.
end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
$endgroup$
First, on $A={0le y<xle 1}$,
begin{align}
mathsf{P}(Xle x,Yle y)&=mathsf{P}(Xle x)-mathsf{P}(Y>y,Xle x) \
&=mathsf{P}left(bigcap_{1le ile n}(U_ile x)right)-mathsf{P}left(bigcap_{1le ile n}(y<U_ile x)right) \
&=x^n-(x-y)^n.
end{align}
Therefore, on $A$,
$$
f_{X,Y}(x,y)=frac{d^2}{dxdy}(x^n-(x-y)^n)=n(n-1)(x-y)^{n-2}.
$$
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
edited Dec 13 '18 at 4:34
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
answered Dec 13 '18 at 4:07
d.k.o.d.k.o.
9,300628
9,300628
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
add a comment |
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
$begingroup$
Thank you so much for your answer. I have a question, though -- would you mind explaining the first and third equalities a little more? I am really not seeing how you arrived at the first one in particular (I'm not questioning if it's correct, I just do not follow!).
$endgroup$
– 0k33
Dec 13 '18 at 4:11
2
2
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
$begingroup$
The event ${X le x}$ can be partitioned into two disjoint events, based on whether $Y le y$ or $Y > y$. The probability of ${X le x}$ is therefore the sum of probabilities of these two smaller events.
$endgroup$
– angryavian
Dec 13 '18 at 4:15
1
1
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
@0k33 The third equality follows from independence and the fact that $mathsf{P}(y<U_ile x)=x-y$.
$endgroup$
– d.k.o.
Dec 13 '18 at 4:36
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
$begingroup$
Thank you both so much! It's clear now!
$endgroup$
– 0k33
Dec 13 '18 at 5:29
add a comment |
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