Record Statistics for Discrete Random Walks
$begingroup$
I code a random walk of length 100 drawn from a LaplaceDistribution:
Accumulate[RandomVariate[LaplaceDistribution[0, 1], 10000]]
I am having trouble counting the number of record events in a discrete random walk. A record occurs at time t if the value of the random walk at time t is greater than all values of the walk for all times less than t. I want a function that will let count the number of records that occur in a walk of length n.
Thanks.
probability-or-statistics random discrete random-process
asked Dec 3 '18 at 22:14
WillWill
3076
$endgroup$
add a comment |
$begingroup$
I code a random walk of length 100 drawn from a LaplaceDistribution:
Accumulate[RandomVariate[LaplaceDistribution[0, 1], 10000]]
I am having trouble counting the number of record events in a discrete random walk. A record occurs at time t if the value of the random walk at time t is greater than all values of the walk for all times less than t. I want a function that will let count the number of records that occur in a walk of length n.
Thanks.
probability-or-statistics random discrete random-process
asked Dec 3 '18 at 22:14
WillWill
3076
$endgroup$
1
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then nBinomial[2 n, n]/2^(2 n - 1)withnthe step number suffices...
$endgroup$
– ciao
Dec 4 '18 at 0:48
add a comment |
$begingroup$
I code a random walk of length 100 drawn from a LaplaceDistribution:
Accumulate[RandomVariate[LaplaceDistribution[0, 1], 10000]]
I am having trouble counting the number of record events in a discrete random walk. A record occurs at time t if the value of the random walk at time t is greater than all values of the walk for all times less than t. I want a function that will let count the number of records that occur in a walk of length n.
Thanks.
probability-or-statistics random discrete random-process
asked Dec 3 '18 at 22:14
WillWill
3076
$endgroup$
I code a random walk of length 100 drawn from a LaplaceDistribution:
Accumulate[RandomVariate[LaplaceDistribution[0, 1], 10000]]
I am having trouble counting the number of record events in a discrete random walk. A record occurs at time t if the value of the random walk at time t is greater than all values of the walk for all times less than t. I want a function that will let count the number of records that occur in a walk of length n.
Thanks.
probability-or-statistics random discrete random-process
probability-or-statistics random discrete random-process
asked Dec 3 '18 at 22:14
WillWill
3076
asked Dec 3 '18 at 22:14
WillWill
3076
edited Dec 4 '18 at 0:51
Will
asked Dec 3 '18 at 22:14
WillWill
3076
asked Dec 3 '18 at 22:14
WillWill
3076
asked Dec 3 '18 at 22:14
WillWill
3076
3076
1
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then nBinomial[2 n, n]/2^(2 n - 1)withnthe step number suffices...
$endgroup$
– ciao
Dec 4 '18 at 0:48
add a comment |
1
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then nBinomial[2 n, n]/2^(2 n - 1)withnthe step number suffices...
$endgroup$
– ciao
Dec 4 '18 at 0:48
1
1
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then n
Binomial[2 n, n]/2^(2 n - 1) with n the step number suffices...$endgroup$
– ciao
Dec 4 '18 at 0:48
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then n
Binomial[2 n, n]/2^(2 n - 1) with n the step number suffices...$endgroup$
– ciao
Dec 4 '18 at 0:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's one take:
path = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 1000]];
records = FoldList[Max, path];
ListPlot[{
path,
records
}]
records has lots of duplicates in it. It's a list where in each position we have the largest value up to that point. If we take the union of the values (or use DeleteDuplicates), we get the unique largest-so-far values, and if we count those we get the desired answer:
Length@Union[records]
52
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
$endgroup$
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
add a comment |
$begingroup$
Let's generate data from a random walk first
SeedRandom[42]
walkdata = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 100]]
, then one way to get what you want is with a Fold:
Last@Fold[
Function[{state,newvalue},
With[{currentrecord=state[[1]],recordcounter=state[[2]]},
If[newvalue > currentrecord,
{newvalue,recordcounter+1},
state
]
]
],
{0,0},
walkdata
]
33
During the fold we keep track of the currentrecord and the number of records (starting with {0,0}) and update it when we find a higher value, otherwise we keep the old. The endresult is the last record and the number of records we encountered from which we just keep the number of record updates (with Last).
Comparing it with C.E.s solution this mainly trades some code clarity (if that's most important definitely go with C.E.s version) for some potential speed up by saving the overhead of the Union function call. If you are dealing with long random walks or doing a lot of them this might become relevant. There is also the additional option to Compile if you need better performance.
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's one take:
path = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 1000]];
records = FoldList[Max, path];
ListPlot[{
path,
records
}]
records has lots of duplicates in it. It's a list where in each position we have the largest value up to that point. If we take the union of the values (or use DeleteDuplicates), we get the unique largest-so-far values, and if we count those we get the desired answer:
Length@Union[records]
52
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
$endgroup$
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
add a comment |
$begingroup$
Here's one take:
path = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 1000]];
records = FoldList[Max, path];
ListPlot[{
path,
records
}]
records has lots of duplicates in it. It's a list where in each position we have the largest value up to that point. If we take the union of the values (or use DeleteDuplicates), we get the unique largest-so-far values, and if we count those we get the desired answer:
Length@Union[records]
52
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
$endgroup$
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
add a comment |
$begingroup$
Here's one take:
path = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 1000]];
records = FoldList[Max, path];
ListPlot[{
path,
records
}]
records has lots of duplicates in it. It's a list where in each position we have the largest value up to that point. If we take the union of the values (or use DeleteDuplicates), we get the unique largest-so-far values, and if we count those we get the desired answer:
Length@Union[records]
52
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
$endgroup$
Here's one take:
path = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 1000]];
records = FoldList[Max, path];
ListPlot[{
path,
records
}]
records has lots of duplicates in it. It's a list where in each position we have the largest value up to that point. If we take the union of the values (or use DeleteDuplicates), we get the unique largest-so-far values, and if we count those we get the desired answer:
Length@Union[records]
52
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
answered Dec 3 '18 at 22:35
C. E.C. E.
50.3k397202
50.3k397202
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
add a comment |
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
Thank you. This is perfect.
$endgroup$
– Will
Dec 3 '18 at 22:51
add a comment |
$begingroup$
Let's generate data from a random walk first
SeedRandom[42]
walkdata = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 100]]
, then one way to get what you want is with a Fold:
Last@Fold[
Function[{state,newvalue},
With[{currentrecord=state[[1]],recordcounter=state[[2]]},
If[newvalue > currentrecord,
{newvalue,recordcounter+1},
state
]
]
],
{0,0},
walkdata
]
33
During the fold we keep track of the currentrecord and the number of records (starting with {0,0}) and update it when we find a higher value, otherwise we keep the old. The endresult is the last record and the number of records we encountered from which we just keep the number of record updates (with Last).
Comparing it with C.E.s solution this mainly trades some code clarity (if that's most important definitely go with C.E.s version) for some potential speed up by saving the overhead of the Union function call. If you are dealing with long random walks or doing a lot of them this might become relevant. There is also the additional option to Compile if you need better performance.
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
$endgroup$
add a comment |
$begingroup$
Let's generate data from a random walk first
SeedRandom[42]
walkdata = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 100]]
, then one way to get what you want is with a Fold:
Last@Fold[
Function[{state,newvalue},
With[{currentrecord=state[[1]],recordcounter=state[[2]]},
If[newvalue > currentrecord,
{newvalue,recordcounter+1},
state
]
]
],
{0,0},
walkdata
]
33
During the fold we keep track of the currentrecord and the number of records (starting with {0,0}) and update it when we find a higher value, otherwise we keep the old. The endresult is the last record and the number of records we encountered from which we just keep the number of record updates (with Last).
Comparing it with C.E.s solution this mainly trades some code clarity (if that's most important definitely go with C.E.s version) for some potential speed up by saving the overhead of the Union function call. If you are dealing with long random walks or doing a lot of them this might become relevant. There is also the additional option to Compile if you need better performance.
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
$endgroup$
add a comment |
$begingroup$
Let's generate data from a random walk first
SeedRandom[42]
walkdata = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 100]]
, then one way to get what you want is with a Fold:
Last@Fold[
Function[{state,newvalue},
With[{currentrecord=state[[1]],recordcounter=state[[2]]},
If[newvalue > currentrecord,
{newvalue,recordcounter+1},
state
]
]
],
{0,0},
walkdata
]
33
During the fold we keep track of the currentrecord and the number of records (starting with {0,0}) and update it when we find a higher value, otherwise we keep the old. The endresult is the last record and the number of records we encountered from which we just keep the number of record updates (with Last).
Comparing it with C.E.s solution this mainly trades some code clarity (if that's most important definitely go with C.E.s version) for some potential speed up by saving the overhead of the Union function call. If you are dealing with long random walks or doing a lot of them this might become relevant. There is also the additional option to Compile if you need better performance.
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
$endgroup$
Let's generate data from a random walk first
SeedRandom[42]
walkdata = Accumulate[RandomVariate[LaplaceDistribution[0, 1], 100]]
, then one way to get what you want is with a Fold:
Last@Fold[
Function[{state,newvalue},
With[{currentrecord=state[[1]],recordcounter=state[[2]]},
If[newvalue > currentrecord,
{newvalue,recordcounter+1},
state
]
]
],
{0,0},
walkdata
]
33
During the fold we keep track of the currentrecord and the number of records (starting with {0,0}) and update it when we find a higher value, otherwise we keep the old. The endresult is the last record and the number of records we encountered from which we just keep the number of record updates (with Last).
Comparing it with C.E.s solution this mainly trades some code clarity (if that's most important definitely go with C.E.s version) for some potential speed up by saving the overhead of the Union function call. If you are dealing with long random walks or doing a lot of them this might become relevant. There is also the additional option to Compile if you need better performance.
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
edited Dec 3 '18 at 23:03
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
answered Dec 3 '18 at 22:37
Thies HeideckeThies Heidecke
6,9362638
6,9362638
add a comment |
add a comment |
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1
$begingroup$
Why do you first generate 10000 samples and then only keep 100 of them randomly?
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:28
$begingroup$
I've realised it is completely pointless to do that lol.
$endgroup$
– Will
Dec 3 '18 at 22:51
$begingroup$
No problem :) Just wanted to clarify if there's some deeper meaning behind it that i missed.
$endgroup$
– Thies Heidecke
Dec 3 '18 at 22:53
$begingroup$
Is the mean number of records in such a walk the desired end result? If so, then n
Binomial[2 n, n]/2^(2 n - 1)withnthe step number suffices...$endgroup$
– ciao
Dec 4 '18 at 0:48