Mixing
How to solve this probability question for the given joint probability distribution? Help mee [closed]
$begingroup$
How can I solve this?
$f_{xy}(x,y)$ is a joint probability distribution defined by
$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And the Question is:
P(min(x,y)<1)=?
Please Write Your Full Answer.
probability
asked Jan 3 at 18:55
Mobina KMobina K
194
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closed as off-topic by amWhy, David K, Did, lulu, metamorphy Jan 4 at 2:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, David K, Did, lulu, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 5 more comments
$begingroup$
How can I solve this?
$f_{xy}(x,y)$ is a joint probability distribution defined by
$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And the Question is:
P(min(x,y)<1)=?
Please Write Your Full Answer.
probability
asked Jan 3 at 18:55
Mobina KMobina K
194
$endgroup$
closed as off-topic by amWhy, David K, Did, lulu, metamorphy Jan 4 at 2:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, David K, Did, lulu, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
2
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
3
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Remarkable—not even a question!
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– LoveTooNap29
Jan 3 at 19:08
2
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Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
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– John Barber
Jan 3 at 19:50
1
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@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07
|
show 5 more comments
$begingroup$
How can I solve this?
$f_{xy}(x,y)$ is a joint probability distribution defined by
$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And the Question is:
P(min(x,y)<1)=?
Please Write Your Full Answer.
probability
asked Jan 3 at 18:55
Mobina KMobina K
194
$endgroup$
How can I solve this?
$f_{xy}(x,y)$ is a joint probability distribution defined by
$$f_{xy}(x,y)=ye^{-y(1+x)}$$ for $ x,y>0$.
And the Question is:
P(min(x,y)<1)=?
Please Write Your Full Answer.
probability
probability
asked Jan 3 at 18:55
Mobina KMobina K
194
asked Jan 3 at 18:55
Mobina KMobina K
194
edited Jan 3 at 20:00
Mobina K
asked Jan 3 at 18:55
Mobina KMobina K
194
asked Jan 3 at 18:55
Mobina KMobina K
194
asked Jan 3 at 18:55
Mobina KMobina K
194
194
closed as off-topic by amWhy, David K, Did, lulu, metamorphy Jan 4 at 2:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, David K, Did, lulu, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, David K, Did, lulu, metamorphy Jan 4 at 2:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, David K, Did, lulu, metamorphy
If this question can be reworded to fit the rules in the help center, please edit the question.
5
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
2
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
3
$begingroup$
Remarkable—not even a question!
$endgroup$
– LoveTooNap29
Jan 3 at 19:08
2
$begingroup$
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
$endgroup$
– John Barber
Jan 3 at 19:50
1
$begingroup$
@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07
|
show 5 more comments
5
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
2
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
3
$begingroup$
Remarkable—not even a question!
$endgroup$
– LoveTooNap29
Jan 3 at 19:08
2
$begingroup$
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
$endgroup$
– John Barber
Jan 3 at 19:50
1
$begingroup$
@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07
5
5
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
2
2
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
3
3
$begingroup$
Remarkable—not even a question!
$endgroup$
– LoveTooNap29
Jan 3 at 19:08
$begingroup$
Remarkable—not even a question!
$endgroup$
– LoveTooNap29
Jan 3 at 19:08
2
2
$begingroup$
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
$endgroup$
– John Barber
Jan 3 at 19:50
$begingroup$
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
$endgroup$
– John Barber
Jan 3 at 19:50
1
1
$begingroup$
@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07
$begingroup$
@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First check normalization. Integrate f(x,y) dx from 0 to infinity to get exp(-y); integrate dy from 0 to infinity to get unity. So, f(x,y) is properly normalized. Now observe condition min(x,y) >= 1 comprises the region [1,infinity) x [1,infinity). P(min(x,y)>=1) is just the measure on this region. Integrate f(x,y) dx from 1 to infinity to get exp(-2y); integrate dy from 1 to infinity to get exp(-2)/2. So P(min(x,y)<1) = 1 - P(min(x,y)>=1) = 1 - exp(-2)/2.
answered Jan 3 at 22:20
WolfWolf
11
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First check normalization. Integrate f(x,y) dx from 0 to infinity to get exp(-y); integrate dy from 0 to infinity to get unity. So, f(x,y) is properly normalized. Now observe condition min(x,y) >= 1 comprises the region [1,infinity) x [1,infinity). P(min(x,y)>=1) is just the measure on this region. Integrate f(x,y) dx from 1 to infinity to get exp(-2y); integrate dy from 1 to infinity to get exp(-2)/2. So P(min(x,y)<1) = 1 - P(min(x,y)>=1) = 1 - exp(-2)/2.
answered Jan 3 at 22:20
WolfWolf
11
$endgroup$
add a comment |
$begingroup$
First check normalization. Integrate f(x,y) dx from 0 to infinity to get exp(-y); integrate dy from 0 to infinity to get unity. So, f(x,y) is properly normalized. Now observe condition min(x,y) >= 1 comprises the region [1,infinity) x [1,infinity). P(min(x,y)>=1) is just the measure on this region. Integrate f(x,y) dx from 1 to infinity to get exp(-2y); integrate dy from 1 to infinity to get exp(-2)/2. So P(min(x,y)<1) = 1 - P(min(x,y)>=1) = 1 - exp(-2)/2.
answered Jan 3 at 22:20
WolfWolf
11
$endgroup$
add a comment |
$begingroup$
First check normalization. Integrate f(x,y) dx from 0 to infinity to get exp(-y); integrate dy from 0 to infinity to get unity. So, f(x,y) is properly normalized. Now observe condition min(x,y) >= 1 comprises the region [1,infinity) x [1,infinity). P(min(x,y)>=1) is just the measure on this region. Integrate f(x,y) dx from 1 to infinity to get exp(-2y); integrate dy from 1 to infinity to get exp(-2)/2. So P(min(x,y)<1) = 1 - P(min(x,y)>=1) = 1 - exp(-2)/2.
answered Jan 3 at 22:20
WolfWolf
11
$endgroup$
First check normalization. Integrate f(x,y) dx from 0 to infinity to get exp(-y); integrate dy from 0 to infinity to get unity. So, f(x,y) is properly normalized. Now observe condition min(x,y) >= 1 comprises the region [1,infinity) x [1,infinity). P(min(x,y)>=1) is just the measure on this region. Integrate f(x,y) dx from 1 to infinity to get exp(-2y); integrate dy from 1 to infinity to get exp(-2)/2. So P(min(x,y)<1) = 1 - P(min(x,y)>=1) = 1 - exp(-2)/2.
answered Jan 3 at 22:20
WolfWolf
11
edited Jan 3 at 22:25
answered Jan 3 at 22:20
WolfWolf
11
answered Jan 3 at 22:20
WolfWolf
11
answered Jan 3 at 22:20
WolfWolf
11
11
add a comment |
add a comment |
5
$begingroup$
I don't see a question here. All you did was to write down a joint probability distribution.
$endgroup$
– lulu
Jan 3 at 18:57
2
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments. Also, you should use MathJax for formatting.
$endgroup$
– saulspatz
Jan 3 at 18:58
3
$begingroup$
Remarkable—not even a question!
$endgroup$
– LoveTooNap29
Jan 3 at 19:08
2
$begingroup$
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
$endgroup$
– John Barber
Jan 3 at 19:50
1
$begingroup$
@HJ_beginner Oh, the $xy$ term in the exponent caught my eye, and I noticed that the marginal pdf of $x$ would have to scale as $1/{(1+x)}^2$. And since $int_0^{infty} dx, x^p / {(1+x)}^2$ diverges for any $pgeq 1$, none of the moments exist.
$endgroup$
– John Barber
Jan 3 at 20:07