Mixing
Markov chain converges to Normal Distribution. How to increase standard deviation
$begingroup$
I am creating an MC (using the following recursive function). It is about a game that each side has different probs of scoring 1 point, 2 points, 3 points or no score (scoreDifference is the score side A - score side B, time is 100 time units, and timeRemaining is what's left.
f[t, d] =
If[t > 100, 0,
If[d == scoreDifference && t == timeRemaining, 1,
p1h*f[t + 1, d - 1] + p2h*f[t + 1, d - 2] + p3h*f[t + 1, d - 3] +
p1a*f[t + 1, d + 1] + p2a*f[t + 1, d + 2] + p3a*f[t + 1, d + 3] +
pns*f[t + 1, d]
]
];
At timeremaining = 0 I get all the possible differences. This converges to a normal distribution (x-axis is the difference, the y-axis is probability).
Example:
p1h = 0.0375 # side A
p2h = 0.05625
p3h = 0.0125
p1a = 0.0327083 # side B
p2a= 0.0490625
p3a = 0.0109028
pns = 0.801076 # noone scores
How can I manipulate this in order to achieve greater (or smaller) standard deviation?
In this example:
E[x] = 11.5
V[x] = 335.6774
SD[x] = 18.3215
normal-distribution markov-chains recursion markov-process
asked Jan 16 at 10:56
TomTom
587
$endgroup$
add a comment |
$begingroup$
I am creating an MC (using the following recursive function). It is about a game that each side has different probs of scoring 1 point, 2 points, 3 points or no score (scoreDifference is the score side A - score side B, time is 100 time units, and timeRemaining is what's left.
f[t, d] =
If[t > 100, 0,
If[d == scoreDifference && t == timeRemaining, 1,
p1h*f[t + 1, d - 1] + p2h*f[t + 1, d - 2] + p3h*f[t + 1, d - 3] +
p1a*f[t + 1, d + 1] + p2a*f[t + 1, d + 2] + p3a*f[t + 1, d + 3] +
pns*f[t + 1, d]
]
];
At timeremaining = 0 I get all the possible differences. This converges to a normal distribution (x-axis is the difference, the y-axis is probability).
Example:
p1h = 0.0375 # side A
p2h = 0.05625
p3h = 0.0125
p1a = 0.0327083 # side B
p2a= 0.0490625
p3a = 0.0109028
pns = 0.801076 # noone scores
How can I manipulate this in order to achieve greater (or smaller) standard deviation?
In this example:
E[x] = 11.5
V[x] = 335.6774
SD[x] = 18.3215
normal-distribution markov-chains recursion markov-process
asked Jan 16 at 10:56
TomTom
587
$endgroup$
$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50
add a comment |
$begingroup$
I am creating an MC (using the following recursive function). It is about a game that each side has different probs of scoring 1 point, 2 points, 3 points or no score (scoreDifference is the score side A - score side B, time is 100 time units, and timeRemaining is what's left.
f[t, d] =
If[t > 100, 0,
If[d == scoreDifference && t == timeRemaining, 1,
p1h*f[t + 1, d - 1] + p2h*f[t + 1, d - 2] + p3h*f[t + 1, d - 3] +
p1a*f[t + 1, d + 1] + p2a*f[t + 1, d + 2] + p3a*f[t + 1, d + 3] +
pns*f[t + 1, d]
]
];
At timeremaining = 0 I get all the possible differences. This converges to a normal distribution (x-axis is the difference, the y-axis is probability).
Example:
p1h = 0.0375 # side A
p2h = 0.05625
p3h = 0.0125
p1a = 0.0327083 # side B
p2a= 0.0490625
p3a = 0.0109028
pns = 0.801076 # noone scores
How can I manipulate this in order to achieve greater (or smaller) standard deviation?
In this example:
E[x] = 11.5
V[x] = 335.6774
SD[x] = 18.3215
normal-distribution markov-chains recursion markov-process
asked Jan 16 at 10:56
TomTom
587
$endgroup$
I am creating an MC (using the following recursive function). It is about a game that each side has different probs of scoring 1 point, 2 points, 3 points or no score (scoreDifference is the score side A - score side B, time is 100 time units, and timeRemaining is what's left.
f[t, d] =
If[t > 100, 0,
If[d == scoreDifference && t == timeRemaining, 1,
p1h*f[t + 1, d - 1] + p2h*f[t + 1, d - 2] + p3h*f[t + 1, d - 3] +
p1a*f[t + 1, d + 1] + p2a*f[t + 1, d + 2] + p3a*f[t + 1, d + 3] +
pns*f[t + 1, d]
]
];
At timeremaining = 0 I get all the possible differences. This converges to a normal distribution (x-axis is the difference, the y-axis is probability).
Example:
p1h = 0.0375 # side A
p2h = 0.05625
p3h = 0.0125
p1a = 0.0327083 # side B
p2a= 0.0490625
p3a = 0.0109028
pns = 0.801076 # noone scores
How can I manipulate this in order to achieve greater (or smaller) standard deviation?
In this example:
E[x] = 11.5
V[x] = 335.6774
SD[x] = 18.3215
normal-distribution markov-chains recursion markov-process
normal-distribution markov-chains recursion markov-process
asked Jan 16 at 10:56
TomTom
587
asked Jan 16 at 10:56
TomTom
587
asked Jan 16 at 10:56
TomTom
587
asked Jan 16 at 10:56
TomTom
587
asked Jan 16 at 10:56
TomTom
587
587
$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50
add a comment |
$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50
$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50
$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50
add a comment |
1 Answer
1
active
oldest
votes
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That is clearly obtained by increasing the probabilities for larger differences ($p_3$'s) wrt those for smaller.
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
$endgroup$
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
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
$begingroup$
That is clearly obtained by increasing the probabilities for larger differences ($p_3$'s) wrt those for smaller.
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
$endgroup$
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
add a comment |
$begingroup$
That is clearly obtained by increasing the probabilities for larger differences ($p_3$'s) wrt those for smaller.
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
$endgroup$
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
add a comment |
$begingroup$
That is clearly obtained by increasing the probabilities for larger differences ($p_3$'s) wrt those for smaller.
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
$endgroup$
That is clearly obtained by increasing the probabilities for larger differences ($p_3$'s) wrt those for smaller.
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
answered Jan 24 at 23:05
G CabG Cab
19.6k31239
19.6k31239
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
add a comment |
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
Unfortunately, I have already tried it. The change is not big enough to get what I want
$endgroup$
– Tom
Jan 25 at 10:03
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
$begingroup$
@Tom: well, if points go from $0$ to $3$, and increasing the prob. for $3$ points does not provide what you want, that only means that the difference in points at each turn is too little. Don't know how you want to alter the rules to cope with that.
$endgroup$
– G Cab
Jan 25 at 11:12
add a comment |
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$begingroup$
Will it help if i convert it to inhomogeneous process? Somehow chamge the probabilities if a player is leading over the other?
$endgroup$
– Tom
Jan 17 at 21:50