How to solve this probability question for the given joint probability distribution? Help mee [on hold]
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 2 days ago
Mobina K
134
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Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by amWhy, David K, Did, lulu, metamorphy 2 days ago
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.
|
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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 2 days ago
Mobina K
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, David K, Did, lulu, metamorphy 2 days ago
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
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
2
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.
– saulspatz
2 days ago
3
Remarkable—not even a question!
– LoveTooNap29
2 days ago
2
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
1
@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.
– John Barber
2 days ago
|
show 5 more comments
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 2 days ago
Mobina K
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Mobina K
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mobina K
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
asked 2 days ago
Mobina K
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mobina K
134
asked 2 days ago
Mobina K
134
134
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mobina K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by amWhy, David K, Did, lulu, metamorphy 2 days ago
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.
put on hold as off-topic by amWhy, David K, Did, lulu, metamorphy 2 days ago
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
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
2
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.
– saulspatz
2 days ago
3
Remarkable—not even a question!
– LoveTooNap29
2 days ago
2
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
1
@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.
– John Barber
2 days ago
|
show 5 more comments
5
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
2
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.
– saulspatz
2 days ago
3
Remarkable—not even a question!
– LoveTooNap29
2 days ago
2
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
1
@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.
– John Barber
2 days ago
5
5
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
2
2
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.
– saulspatz
2 days ago
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.
– saulspatz
2 days ago
3
3
Remarkable—not even a question!
– LoveTooNap29
2 days ago
Remarkable—not even a question!
– LoveTooNap29
2 days ago
2
2
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
1
1
@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.
– John Barber
2 days ago
@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.
– John Barber
2 days ago
|
show 5 more comments
1 Answer
1
active
oldest
votes
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 2 days ago
Wolf
11
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Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
Wolf
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 2 days ago
Wolf
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 2 days ago
Wolf
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
Wolf
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
answered 2 days ago
Wolf
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 days ago
Wolf
11
answered 2 days ago
Wolf
11
11
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wolf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
5
I don't see a question here. All you did was to write down a joint probability distribution.
– lulu
2 days ago
2
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.
– saulspatz
2 days ago
3
Remarkable—not even a question!
– LoveTooNap29
2 days ago
2
Well, for whatever it's worth, this joint pdf is somewhat interesting because $x$ has no finite moments.
– John Barber
2 days ago
1
@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.
– John Barber
2 days ago