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 10 hours ago
gwr
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 10 hours ago
gwr
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 10 hours ago
gwr
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 10 hours ago
gwr
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 10 hours ago
gwr
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 10 hours ago
gwr
7,63322558
edited 10 hours ago
gwr
7,63322558
edited 10 hours ago
gwr
7,63322558
7,63322558
asked 13 hours ago
Abdul HaqAbdul Haq
191
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 13 hours ago
Abdul HaqAbdul Haq
191
asked 13 hours ago
Abdul HaqAbdul Haq
191
191
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 11 hours 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 11 hours 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 11 hours 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 11 hours 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 11 hours ago
Bob HanlonBob Hanlon
59.2k33595
answered 11 hours ago
Bob HanlonBob Hanlon
59.2k33595
answered 11 hours ago
Bob HanlonBob Hanlon
59.2k33595
answered 11 hours 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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