Problem with the Inverse CDF of Non-central F Ratio Distribution
$begingroup$
In[3]:= n = 5; n1 = 4; n2 = 6; γ = 0.05; α = 1/370;
InverseCDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2],
1 - α - (n - n1)/n2]
During evaluation of In[3]:= FindRoot::lstol: The line search
decreased the step size to within tolerance specified by AccuracyGoal
and PrecisionGoal but was unable to find a sufficient decrease in the
merit function. You may need more than MachinePrecision digits of
working precision to meet these tolerances. >>
Out[4]= 3.24659
The correct answer is 5468.146427955789
numerics probability-or-statistics accuracy-and-precision
edited 1 hour ago
gwr
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In[3]:= n = 5; n1 = 4; n2 = 6; γ = 0.05; α = 1/370;
InverseCDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2],
1 - α - (n - n1)/n2]
During evaluation of In[3]:= FindRoot::lstol: The line search
decreased the step size to within tolerance specified by AccuracyGoal
and PrecisionGoal but was unable to find a sufficient decrease in the
merit function. You may need more than MachinePrecision digits of
working precision to meet these tolerances. >>
Out[4]= 3.24659
The correct answer is 5468.146427955789
numerics probability-or-statistics accuracy-and-precision
edited 1 hour ago
gwr
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In[3]:= n = 5; n1 = 4; n2 = 6; γ = 0.05; α = 1/370;
InverseCDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2],
1 - α - (n - n1)/n2]
During evaluation of In[3]:= FindRoot::lstol: The line search
decreased the step size to within tolerance specified by AccuracyGoal
and PrecisionGoal but was unable to find a sufficient decrease in the
merit function. You may need more than MachinePrecision digits of
working precision to meet these tolerances. >>
Out[4]= 3.24659
The correct answer is 5468.146427955789
numerics probability-or-statistics accuracy-and-precision
edited 1 hour ago
gwr
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In[3]:= n = 5; n1 = 4; n2 = 6; γ = 0.05; α = 1/370;
InverseCDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2],
1 - α - (n - n1)/n2]
During evaluation of In[3]:= FindRoot::lstol: The line search
decreased the step size to within tolerance specified by AccuracyGoal
and PrecisionGoal but was unable to find a sufficient decrease in the
merit function. You may need more than MachinePrecision digits of
working precision to meet these tolerances. >>
Out[4]= 3.24659
The correct answer is 5468.146427955789
numerics probability-or-statistics accuracy-and-precision
numerics probability-or-statistics accuracy-and-precision
edited 1 hour ago
gwr
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
gwr
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
gwr
7,58822558
edited 1 hour ago
gwr
7,58822558
edited 1 hour ago
gwr
7,58822558
7,58822558
asked 4 hours ago
Abdul HaqAbdul Haq
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Abdul HaqAbdul Haq
141
asked 4 hours ago
Abdul HaqAbdul Haq
141
141
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Abdul Haq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Abdul Haq 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 |
1 Answer
1
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$begingroup$
Use FindRoot directly with arbitrary-precision
n = 5; n1 = 4; n2 = 6; γ = 1/20; α = 1/370;
icdf = x /.
FindRoot[CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], x] ==
1 - α - (n - n1)/n2, {x, 500},
WorkingPrecision -> $MachinePrecision]
(* 5468.146403807255 *)
Verifying,
(CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], icdf] //
RootApproximant) === 1 - α - (n - n1)/n2
(* True *)
Note that while the starting value used in FindRoot needs to be large, it does not have to be particularly close to the actual value.
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use FindRoot directly with arbitrary-precision
n = 5; n1 = 4; n2 = 6; γ = 1/20; α = 1/370;
icdf = x /.
FindRoot[CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], x] ==
1 - α - (n - n1)/n2, {x, 500},
WorkingPrecision -> $MachinePrecision]
(* 5468.146403807255 *)
Verifying,
(CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], icdf] //
RootApproximant) === 1 - α - (n - n1)/n2
(* True *)
Note that while the starting value used in FindRoot needs to be large, it does not have to be particularly close to the actual value.
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
$endgroup$
add a comment |
$begingroup$
Use FindRoot directly with arbitrary-precision
n = 5; n1 = 4; n2 = 6; γ = 1/20; α = 1/370;
icdf = x /.
FindRoot[CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], x] ==
1 - α - (n - n1)/n2, {x, 500},
WorkingPrecision -> $MachinePrecision]
(* 5468.146403807255 *)
Verifying,
(CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], icdf] //
RootApproximant) === 1 - α - (n - n1)/n2
(* True *)
Note that while the starting value used in FindRoot needs to be large, it does not have to be particularly close to the actual value.
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
$endgroup$
add a comment |
$begingroup$
Use FindRoot directly with arbitrary-precision
n = 5; n1 = 4; n2 = 6; γ = 1/20; α = 1/370;
icdf = x /.
FindRoot[CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], x] ==
1 - α - (n - n1)/n2, {x, 500},
WorkingPrecision -> $MachinePrecision]
(* 5468.146403807255 *)
Verifying,
(CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], icdf] //
RootApproximant) === 1 - α - (n - n1)/n2
(* True *)
Note that while the starting value used in FindRoot needs to be large, it does not have to be particularly close to the actual value.
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
$endgroup$
Use FindRoot directly with arbitrary-precision
n = 5; n1 = 4; n2 = 6; γ = 1/20; α = 1/370;
icdf = x /.
FindRoot[CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], x] ==
1 - α - (n - n1)/n2, {x, 500},
WorkingPrecision -> $MachinePrecision]
(* 5468.146403807255 *)
Verifying,
(CDF[NoncentralFRatioDistribution[1, n1 - 1, n1/γ^2], icdf] //
RootApproximant) === 1 - α - (n - n1)/n2
(* True *)
Note that while the starting value used in FindRoot needs to be large, it does not have to be particularly close to the actual value.
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
answered 1 hour ago
Bob HanlonBob Hanlon
59.2k33595
59.2k33595
add a comment |
add a comment |
Abdul Haq is a new contributor. Be nice, and check out our Code of Conduct.
Abdul Haq is a new contributor. Be nice, and check out our Code of Conduct.
Abdul Haq is a new contributor. Be nice, and check out our Code of Conduct.
Abdul Haq is a new contributor. Be nice, and check out our Code of Conduct.
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