Mixing
Find $P(A>B+C)$ if A,B,C are independent and uniformly distributed in [0,1]
$begingroup$
Considering this I have tried:
$$ P(A>B+C)=P(C<A-B)=int_0^1int_0^1 P(C<a-b|A=a, B=b),da,db =$$
$$=int_0^1int_0^1 P(C<a-b),f_A(a),f_b(b),da,db = int_0^1int_0^1 F_C(a-b),f_A(a),f_B(b),da,db=$$
$$=int_0^1int_0^1 (a-b) cdot 1 cdot 1 ,da,db= 0$$
The actual answer is $frac{1}{6}$.
I feel the mistake is to eliminate both A and B at the same time. Unfortunately I can't figure out the right solution. Any help will be higly appreciated!
probability
asked Jan 26 at 16:32
gerrokgerrok
83
$endgroup$
add a comment |
$begingroup$
Considering this I have tried:
$$ P(A>B+C)=P(C<A-B)=int_0^1int_0^1 P(C<a-b|A=a, B=b),da,db =$$
$$=int_0^1int_0^1 P(C<a-b),f_A(a),f_b(b),da,db = int_0^1int_0^1 F_C(a-b),f_A(a),f_B(b),da,db=$$
$$=int_0^1int_0^1 (a-b) cdot 1 cdot 1 ,da,db= 0$$
The actual answer is $frac{1}{6}$.
I feel the mistake is to eliminate both A and B at the same time. Unfortunately I can't figure out the right solution. Any help will be higly appreciated!
probability
asked Jan 26 at 16:32
gerrokgerrok
83
$endgroup$
add a comment |
$begingroup$
Considering this I have tried:
$$ P(A>B+C)=P(C<A-B)=int_0^1int_0^1 P(C<a-b|A=a, B=b),da,db =$$
$$=int_0^1int_0^1 P(C<a-b),f_A(a),f_b(b),da,db = int_0^1int_0^1 F_C(a-b),f_A(a),f_B(b),da,db=$$
$$=int_0^1int_0^1 (a-b) cdot 1 cdot 1 ,da,db= 0$$
The actual answer is $frac{1}{6}$.
I feel the mistake is to eliminate both A and B at the same time. Unfortunately I can't figure out the right solution. Any help will be higly appreciated!
probability
asked Jan 26 at 16:32
gerrokgerrok
83
$endgroup$
Considering this I have tried:
$$ P(A>B+C)=P(C<A-B)=int_0^1int_0^1 P(C<a-b|A=a, B=b),da,db =$$
$$=int_0^1int_0^1 P(C<a-b),f_A(a),f_b(b),da,db = int_0^1int_0^1 F_C(a-b),f_A(a),f_B(b),da,db=$$
$$=int_0^1int_0^1 (a-b) cdot 1 cdot 1 ,da,db= 0$$
The actual answer is $frac{1}{6}$.
I feel the mistake is to eliminate both A and B at the same time. Unfortunately I can't figure out the right solution. Any help will be higly appreciated!
probability
probability
asked Jan 26 at 16:32
gerrokgerrok
83
asked Jan 26 at 16:32
gerrokgerrok
83
asked Jan 26 at 16:32
gerrokgerrok
83
asked Jan 26 at 16:32
gerrokgerrok
83
asked Jan 26 at 16:32
gerrokgerrok
83
83
add a comment |
add a comment |
2 Answers
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$begingroup$
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) in [0,1]times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
add a comment |
$begingroup$
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) in [0,1]times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
add a comment |
$begingroup$
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) in [0,1]times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
$endgroup$
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
add a comment |
$begingroup$
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) in [0,1]times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
$endgroup$
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) in [0,1]times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
answered Jan 26 at 16:51
inhuretnakhtinhuretnakht
38617
38617
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
add a comment |
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
$begingroup$
Got it. Will try to figure out the right limits. Thank you!
$endgroup$
– gerrok
Jan 26 at 17:11
add a comment |
$begingroup$
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
$endgroup$
add a comment |
$begingroup$
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
$endgroup$
add a comment |
$begingroup$
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
$endgroup$
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
answered Jan 26 at 17:12
kimchi loverkimchi lover
11.4k31229
11.4k31229
add a comment |
add a comment |
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