Problems on Normal Variable [closed]

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I have a question on this problem：




You have a receiver that can receive two signals, signal $A$ and signal $B$. Both signals have $0$ mean and typically arrive with same frequency. Signal $A$ is normally distributed with variance $4$ and $B$ is normally distributed with variance $9$. You observe a signal with magnitude $2$. What is the probability the signal you observe is signal $A$?







probability







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closed as off-topic by heropup, user99914, Adrian Keister, Jendrik Stelzner, Xander Henderson Sep 8 at 1:48


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, Adrian Keister, Jendrik Stelzner, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Before you measured the the signal, were the two sources equally likely?
    – Henry
    Sep 7 at 10:50










  • I think they are just independent random variables
    – Z-Harlpet
    Sep 7 at 22:40














up vote
1
down vote

favorite












I have a question on this problem：




You have a receiver that can receive two signals, signal $A$ and signal $B$. Both signals have $0$ mean and typically arrive with same frequency. Signal $A$ is normally distributed with variance $4$ and $B$ is normally distributed with variance $9$. You observe a signal with magnitude $2$. What is the probability the signal you observe is signal $A$?







probability







share|cite|improve this question















closed as off-topic by heropup, user99914, Adrian Keister, Jendrik Stelzner, Xander Henderson Sep 8 at 1:48


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, Adrian Keister, Jendrik Stelzner, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Before you measured the the signal, were the two sources equally likely?
    – Henry
    Sep 7 at 10:50










  • I think they are just independent random variables
    – Z-Harlpet
    Sep 7 at 22:40












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have a question on this problem：




You have a receiver that can receive two signals, signal $A$ and signal $B$. Both signals have $0$ mean and typically arrive with same frequency. Signal $A$ is normally distributed with variance $4$ and $B$ is normally distributed with variance $9$. You observe a signal with magnitude $2$. What is the probability the signal you observe is signal $A$?







probability







share|cite|improve this question















I have a question on this problem：




You have a receiver that can receive two signals, signal $A$ and signal $B$. Both signals have $0$ mean and typically arrive with same frequency. Signal $A$ is normally distributed with variance $4$ and $B$ is normally distributed with variance $9$. You observe a signal with magnitude $2$. What is the probability the signal you observe is signal $A$?







probability




probability






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share|cite|improve this question













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share|cite|improve this question








edited Sep 8 at 0:28









Jendrik Stelzner

7,69121137




7,69121137










asked Sep 7 at 9:15









Z-Harlpet

307




307




closed as off-topic by heropup, user99914, Adrian Keister, Jendrik Stelzner, Xander Henderson Sep 8 at 1:48


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, Adrian Keister, Jendrik Stelzner, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by heropup, user99914, Adrian Keister, Jendrik Stelzner, Xander Henderson Sep 8 at 1:48


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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, Adrian Keister, Jendrik Stelzner, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    Before you measured the the signal, were the two sources equally likely?
    – Henry
    Sep 7 at 10:50










  • I think they are just independent random variables
    – Z-Harlpet
    Sep 7 at 22:40












  • 2




    Before you measured the the signal, were the two sources equally likely?
    – Henry
    Sep 7 at 10:50










  • I think they are just independent random variables
    – Z-Harlpet
    Sep 7 at 22:40







2




2




Before you measured the the signal, were the two sources equally likely?
– Henry
Sep 7 at 10:50




Before you measured the the signal, were the two sources equally likely?
– Henry
Sep 7 at 10:50












I think they are just independent random variables
– Z-Harlpet
Sep 7 at 22:40




I think they are just independent random variables
– Z-Harlpet
Sep 7 at 22:40










1 Answer
1






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oldest

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up vote
0
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The probability density for $N(0,sigma)$ is $p_sigma(x)=frac1sqrt2pisigmae^frac-x^22sigma^2$. Therefore signal probabiliity is $A$ is$P(A)=fracp_2(2)p_2(2)+p_3(2)$.






share|cite|improve this answer




















  • I think we should consider conditional probability.
    – Z-Harlpet
    Sep 7 at 22:41










  • What conditions?
    – herb steinberg
    Sep 8 at 1:11

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













The probability density for $N(0,sigma)$ is $p_sigma(x)=frac1sqrt2pisigmae^frac-x^22sigma^2$. Therefore signal probabiliity is $A$ is$P(A)=fracp_2(2)p_2(2)+p_3(2)$.






share|cite|improve this answer




















  • I think we should consider conditional probability.
    – Z-Harlpet
    Sep 7 at 22:41










  • What conditions?
    – herb steinberg
    Sep 8 at 1:11














up vote
0
down vote













The probability density for $N(0,sigma)$ is $p_sigma(x)=frac1sqrt2pisigmae^frac-x^22sigma^2$. Therefore signal probabiliity is $A$ is$P(A)=fracp_2(2)p_2(2)+p_3(2)$.






share|cite|improve this answer




















  • I think we should consider conditional probability.
    – Z-Harlpet
    Sep 7 at 22:41










  • What conditions?
    – herb steinberg
    Sep 8 at 1:11












up vote
0
down vote










up vote
0
down vote









The probability density for $N(0,sigma)$ is $p_sigma(x)=frac1sqrt2pisigmae^frac-x^22sigma^2$. Therefore signal probabiliity is $A$ is$P(A)=fracp_2(2)p_2(2)+p_3(2)$.






share|cite|improve this answer












The probability density for $N(0,sigma)$ is $p_sigma(x)=frac1sqrt2pisigmae^frac-x^22sigma^2$. Therefore signal probabiliity is $A$ is$P(A)=fracp_2(2)p_2(2)+p_3(2)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 7 at 16:22









herb steinberg

1,476210




1,476210











  • I think we should consider conditional probability.
    – Z-Harlpet
    Sep 7 at 22:41










  • What conditions?
    – herb steinberg
    Sep 8 at 1:11
















  • I think we should consider conditional probability.
    – Z-Harlpet
    Sep 7 at 22:41










  • What conditions?
    – herb steinberg
    Sep 8 at 1:11















I think we should consider conditional probability.
– Z-Harlpet
Sep 7 at 22:41




I think we should consider conditional probability.
– Z-Harlpet
Sep 7 at 22:41












What conditions?
– herb steinberg
Sep 8 at 1:11




What conditions?
– herb steinberg
Sep 8 at 1:11


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