Find the probability (normal + exponential)
$begingroup$
Let $X$ be normal with mean $0$ and variance $1/2$, Y exponential with mean $1$. $X$ and $Y$ are independent. Find $mathbb{P}(Y>X^2)$
$mathbb{P}(Y>X^2)=intlimits_{-infty}^{infty}mathbb{P}(Y>X^2|X=x)f_{X}(x)dx= ...=sqrt{2}/2$
Second approach. Since $2X^2$ is $Gamma(1/2,1/2)$, then $2cdot2X^2$ is $Gamma(1,1/2)$ which is exponential with mean $2$. Thus $mathbb{P}(Y>X^2)=P(4Y>4X^2)=P(Y>0.25T)=aneqfrac{sqrt{2}}{2}$, where $T$ is exponentially distributed, indepedent of $Y$.
The question is: where am I wrong? I suspect the second approach is wrong, but why?
probability
asked Jun 8 '13 at 20:10
czachurczachur
1769
$endgroup$
add a comment |
$begingroup$
Let $X$ be normal with mean $0$ and variance $1/2$, Y exponential with mean $1$. $X$ and $Y$ are independent. Find $mathbb{P}(Y>X^2)$
$mathbb{P}(Y>X^2)=intlimits_{-infty}^{infty}mathbb{P}(Y>X^2|X=x)f_{X}(x)dx= ...=sqrt{2}/2$
Second approach. Since $2X^2$ is $Gamma(1/2,1/2)$, then $2cdot2X^2$ is $Gamma(1,1/2)$ which is exponential with mean $2$. Thus $mathbb{P}(Y>X^2)=P(4Y>4X^2)=P(Y>0.25T)=aneqfrac{sqrt{2}}{2}$, where $T$ is exponentially distributed, indepedent of $Y$.
The question is: where am I wrong? I suspect the second approach is wrong, but why?
probability
asked Jun 8 '13 at 20:10
czachurczachur
1769
$endgroup$
1
$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21
add a comment |
$begingroup$
Let $X$ be normal with mean $0$ and variance $1/2$, Y exponential with mean $1$. $X$ and $Y$ are independent. Find $mathbb{P}(Y>X^2)$
$mathbb{P}(Y>X^2)=intlimits_{-infty}^{infty}mathbb{P}(Y>X^2|X=x)f_{X}(x)dx= ...=sqrt{2}/2$
Second approach. Since $2X^2$ is $Gamma(1/2,1/2)$, then $2cdot2X^2$ is $Gamma(1,1/2)$ which is exponential with mean $2$. Thus $mathbb{P}(Y>X^2)=P(4Y>4X^2)=P(Y>0.25T)=aneqfrac{sqrt{2}}{2}$, where $T$ is exponentially distributed, indepedent of $Y$.
The question is: where am I wrong? I suspect the second approach is wrong, but why?
probability
asked Jun 8 '13 at 20:10
czachurczachur
1769
$endgroup$
Let $X$ be normal with mean $0$ and variance $1/2$, Y exponential with mean $1$. $X$ and $Y$ are independent. Find $mathbb{P}(Y>X^2)$
$mathbb{P}(Y>X^2)=intlimits_{-infty}^{infty}mathbb{P}(Y>X^2|X=x)f_{X}(x)dx= ...=sqrt{2}/2$
Second approach. Since $2X^2$ is $Gamma(1/2,1/2)$, then $2cdot2X^2$ is $Gamma(1,1/2)$ which is exponential with mean $2$. Thus $mathbb{P}(Y>X^2)=P(4Y>4X^2)=P(Y>0.25T)=aneqfrac{sqrt{2}}{2}$, where $T$ is exponentially distributed, indepedent of $Y$.
The question is: where am I wrong? I suspect the second approach is wrong, but why?
probability
probability
asked Jun 8 '13 at 20:10
czachurczachur
1769
asked Jun 8 '13 at 20:10
czachurczachur
1769
asked Jun 8 '13 at 20:10
czachurczachur
1769
asked Jun 8 '13 at 20:10
czachurczachur
1769
asked Jun 8 '13 at 20:10
czachurczachur
1769
1769
1
$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21
add a comment |
1
$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21
1
1
$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21
$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
More generally: given $X$~$N(mu,sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:
Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:
Then, for your given parameter values, the correct solution is:
which is approximately 0.7.
Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $mu=0$ and $sigma = sqrtfrac12$):
datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000];
datay = RandomReal[ExponentialDistribution[1], 100000];
Count[datay - datax^2, x_ /; x > 0] /100000.
0.70418
:)
OP wrote: - Second approach. "Since $2X^2$ is $Gamma(1/2,1/2)$ ..."
Correctly: Since $sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(frac12,2)$ ... not Gamma$(frac12,frac12)$.
edited Jan 17 at 21:36
Glorfindel
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
$endgroup$
$begingroup$
Quote:The question is: where am I wrong?
$endgroup$
– Did
Jun 9 '13 at 16:18
2
$begingroup$
@DidThe answer is: everywhere
$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
$begingroup$
@Did The OP's question title is:Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.
$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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$begingroup$
More generally: given $X$~$N(mu,sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:
Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:
Then, for your given parameter values, the correct solution is:
which is approximately 0.7.
Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $mu=0$ and $sigma = sqrtfrac12$):
datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000];
datay = RandomReal[ExponentialDistribution[1], 100000];
Count[datay - datax^2, x_ /; x > 0] /100000.
0.70418
:)
OP wrote: - Second approach. "Since $2X^2$ is $Gamma(1/2,1/2)$ ..."
Correctly: Since $sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(frac12,2)$ ... not Gamma$(frac12,frac12)$.
edited Jan 17 at 21:36
Glorfindel
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
$endgroup$
$begingroup$
Quote:The question is: where am I wrong?
$endgroup$
– Did
Jun 9 '13 at 16:18
2
$begingroup$
@DidThe answer is: everywhere
$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
$begingroup$
@Did The OP's question title is:Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.
$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
|
show 1 more comment
$begingroup$
More generally: given $X$~$N(mu,sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:
Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:
Then, for your given parameter values, the correct solution is:
which is approximately 0.7.
Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $mu=0$ and $sigma = sqrtfrac12$):
datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000];
datay = RandomReal[ExponentialDistribution[1], 100000];
Count[datay - datax^2, x_ /; x > 0] /100000.
0.70418
:)
OP wrote: - Second approach. "Since $2X^2$ is $Gamma(1/2,1/2)$ ..."
Correctly: Since $sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(frac12,2)$ ... not Gamma$(frac12,frac12)$.
edited Jan 17 at 21:36
Glorfindel
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
$endgroup$
$begingroup$
Quote:The question is: where am I wrong?
$endgroup$
– Did
Jun 9 '13 at 16:18
2
$begingroup$
@DidThe answer is: everywhere
$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
$begingroup$
@Did The OP's question title is:Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.
$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
|
show 1 more comment
$begingroup$
More generally: given $X$~$N(mu,sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:
Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:
Then, for your given parameter values, the correct solution is:
which is approximately 0.7.
Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $mu=0$ and $sigma = sqrtfrac12$):
datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000];
datay = RandomReal[ExponentialDistribution[1], 100000];
Count[datay - datax^2, x_ /; x > 0] /100000.
0.70418
:)
OP wrote: - Second approach. "Since $2X^2$ is $Gamma(1/2,1/2)$ ..."
Correctly: Since $sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(frac12,2)$ ... not Gamma$(frac12,frac12)$.
edited Jan 17 at 21:36
Glorfindel
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
$endgroup$
More generally: given $X$~$N(mu,sigma^2)$ and $Y$~Exponential(1), with $X$ and $Y$ independent, their joint pdf is $f(x,y)$:
Here is a quick check of the probability you seek, using the mathStatica add-on to Mathematica:
Then, for your given parameter values, the correct solution is:
which is approximately 0.7.
Finally, a quick Monte Carlo check to confirm we haven't made any mistakes ... (the following sets $mu=0$ and $sigma = sqrtfrac12$):
datax = RandomReal[NormalDistribution[0, Sqrt[1/2]], 100000];
datay = RandomReal[ExponentialDistribution[1], 100000];
Count[datay - datax^2, x_ /; x > 0] /100000.
0.70418
:)
OP wrote: - Second approach. "Since $2X^2$ is $Gamma(1/2,1/2)$ ..."
Correctly: Since $sqrt{2}X$ ~ $N(0,1)$, and the square of a standard Normal is Chisquared(1), it follows that: $2X^2$ ~ Chisquared(1) ... which is: Gamma$(frac12,2)$ ... not Gamma$(frac12,frac12)$.
edited Jan 17 at 21:36
Glorfindel
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
edited Jan 17 at 21:36
Glorfindel
3,41381930
edited Jan 17 at 21:36
Glorfindel
3,41381930
edited Jan 17 at 21:36
Glorfindel
3,41381930
3,41381930
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
answered Jun 9 '13 at 6:26
wolfieswolfies
4,2692923
4,2692923
$begingroup$
Quote:The question is: where am I wrong?
$endgroup$
– Did
Jun 9 '13 at 16:18
2
$begingroup$
@DidThe answer is: everywhere
$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
$begingroup$
@Did The OP's question title is:Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.
$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
|
show 1 more comment
$begingroup$
Quote:The question is: where am I wrong?
$endgroup$
– Did
Jun 9 '13 at 16:18
2
$begingroup$
@DidThe answer is: everywhere
$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
$begingroup$
@Did The OP's question title is:Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.
$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
$begingroup$
Quote:
The question is: where am I wrong?$endgroup$
– Did
Jun 9 '13 at 16:18
$begingroup$
Quote:
The question is: where am I wrong?$endgroup$
– Did
Jun 9 '13 at 16:18
2
2
$begingroup$
@Did
The answer is: everywhere$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
@Did
The answer is: everywhere$endgroup$
– wolfies
Jun 9 '13 at 16:50
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
$begingroup$
Then, how is your post describing/identifying/remedying to, this "everywhere"? That is, how is it answering the question?
$endgroup$
– Did
Jun 9 '13 at 17:37
1
1
$begingroup$
@Did The OP's question title is:
Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.$endgroup$
– wolfies
Jun 9 '13 at 17:57
$begingroup$
@Did The OP's question title is:
Find the probability. This I have done. If you would like to address his second-level question as to how/where he went wrong, the site does allow (and in fact encourages) others to submit solutions that may be helpful, and I would certainly encourage you to do so. Though, as this may possibly be a homework assignment, you might want to take care not to unduly interfere with the parent university's ability to carry out proper assessments.$endgroup$
– wolfies
Jun 9 '13 at 17:57
1
1
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
$begingroup$
As @Did probably knows, displaying a CAS calculation and Monte Carlo simulation that agree with the OP's first calculation and disagree with the second, answers part of "where am I wrong" and confirms the "I suspect the second approach is wrong". It is wasting everyone's time to post a potentially endless sequence of negative comments in which one user tries to control the interpretation of words like question and answer in a quixotic campaign to demonstrate that this (useful) answer and ones like it are not a valid response to the OP's question, or are "not mathematical".
$endgroup$
– zyx
Jun 10 '13 at 3:13
|
show 1 more comment
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$begingroup$
$2*2X^2simGamma(1/2,1)$ if $2X^2simGamma(1/2,1/2)$
$endgroup$
– Seyhmus Güngören
Jun 8 '13 at 20:21