Find the expected number of stages until one of the players is eliminated.
1
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I have been given an example:
Moreover I have been given a problem based on this example: 
I tried to find an appropriate martingale to solve this problem but I couldn't. I would appreciate any tips and hints.
probability stochastic-processes martingales
asked Jan 10 at 20:51
WywanaWywana
685
$endgroup$
add a comment |
1
$begingroup$
I have been given an example:
Moreover I have been given a problem based on this example:
I tried to find an appropriate martingale to solve this problem but I couldn't. I would appreciate any tips and hints.
probability stochastic-processes martingales
asked Jan 10 at 20:51
WywanaWywana
685
$endgroup$
$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35
add a comment |
1
1
1
1
$begingroup$
I have been given an example:
Moreover I have been given a problem based on this example:
I tried to find an appropriate martingale to solve this problem but I couldn't. I would appreciate any tips and hints.
probability stochastic-processes martingales
asked Jan 10 at 20:51
WywanaWywana
685
$endgroup$
I have been given an example:
Moreover I have been given a problem based on this example:
I tried to find an appropriate martingale to solve this problem but I couldn't. I would appreciate any tips and hints.
probability stochastic-processes martingales
probability stochastic-processes martingales
asked Jan 10 at 20:51
WywanaWywana
685
asked Jan 10 at 20:51
WywanaWywana
685
edited Jan 10 at 22:32
Wywana
asked Jan 10 at 20:51
WywanaWywana
685
asked Jan 10 at 20:51
WywanaWywana
685
asked Jan 10 at 20:51
WywanaWywana
685
685
$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35
add a comment |
$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35
$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35
add a comment |
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$begingroup$
The two player game is known to have an expected length that is the product of the starting numbers of coins.
$endgroup$
– Ross Millikan
Jan 10 at 22:40
$begingroup$
@RossMillikan I have thought about $ M_n = X_n Y_n Z_n $, unfortunately it's not a martingale $ E(M_{n+1} | X_n = X, Y_n=Y, Z_n = Z) = XYZ -frac{1}{3}(X + Y + Z) $.
$endgroup$
– Wywana
Jan 11 at 0:35