by Kevin Harlow
July 21, 2003
Problem: Estimate BB, S, D, T, HR based on AB, BA, OBP, SLG
There are 5 unknowns (BB, S, D, T, and HR) and only three equations (BA, OBP, SLG). We must find empirical relationships for two more equations. Using Lahman's database, each player-season with a minimum of 300 PA since 1920 (lively ball era) is considered. A regression of each event rate versus (BA, OBP, SLG) was performed. Based upon the goodness of fit results given in Table 1, S/PA and HR/PA can be estimated well using (BA, OBP, SLG).
Event Rate -- R^2
S/PA -- 0.941
D/PA -- 0.321
T/PA -- 0.124
HR/PA -- 0.918
Specifically, the two empirical equations are:
S = (0.03298 + 1.341779*AVG - 0.24882*OBP - 0.33932*SLG) * (AB+BB)
HR = (0.011458 - 0.35722*AVG - 0.02256*OBP + 0.281748*SLG) * (AB+BB)
These two empirical equations combined with the three definitions below
BA = (S+D+T+HR) / (AB)
OBP = (S+D+T+HR+BB) / (AB+BB)
SLG = [S+(2*D)+(3*T)+(4*HR)] / (AB)
consist of the necessary 5 equations needed to solve for the five unknowns.
In order to solve these equations I first rearrange the three definitions such that the RHS of each rearranged equation is equal to the same thing.
AB = (S+D+T+HR) / (BA)
AB = [S+D+T+HR+BB*(1-OBP)] / (OBP)
AB = [S+(2*D)+(3*T)+(4*HR)] / (SLG)
BB is solved for first by setting the BA and OBP equations equal.
(S+D+T+HR)/(BA) = (S+D+T+HR)/(OBP) + (BB*(1-OBP))/(OBP)
(1/BA - 1/OBP)*(S+D+T+HR) = (BB*(1-OBP))/(OBP)
(S+D+T+HR) = BB * ((1-OBP)*BA/(OBP-BA))
Substituting this into the rearranged BA equation
AB = (S+D+T+HR)/(BA)
AB = BB * [(1-OBP)*BA] / [(OBP-BA)*BA]
BB = (OBP-BA)/(1-OBP) * AB
Now, set the rearranged BA and SLG equations equal.
(S+D+T+HR)/(BA) = [S+(2*D)+(3*T)+(4*HR)]/(SLG)
Even though an empirical equation for S has already been determined, solve the above equation in terms of S.
(1/BA-1/SLG)*S = (2/SLG-1/BA)*D + (3/SLG-1/BA)*T + (4/SLG-1/BA)*HR
S = ((2*BA-SLG)/(SLG-BA))*D + ((3*BA-SLG)/(SLG-BA))*T + ((4*BA-SLG)/(SLG-BA))*HR
Define the following:
X1 = (2*BA-SLG) / (SLG-BA)
X2 = (3*BA-SLG) / (SLG-BA)
X3 = (4*BA-SLG) / (SLG-BA)
S = X1*D + X2*T + X3*HR
Substitute this derived equation for S into the rearranged SLG and OBP equations.
AB = [(X1+2)*D + (X2+3)*T + (X3+4)*HR] / (SLG)
AB = [(X1+1)*D + (X2+1)*T + (X3+1)*HR + BB*(1-OBP)] / (OBP)
Setting these two equations equal and solving for D gives
[(X1+2)/SLG-(X1+1)/OBP]*D = [(X3+1)/OBP-(X3+4)/SLG]*HR + [(X2+1)/OBP-(X2+3)/SLG]*T + [(1-OBP)/OBP]*BB
Define the following:
Y1 = [(X3+1)*SLG-(X3+4)*OBP] / [(X1+2)*OBP-(X1+1)*SLG]
Y2 = [(X2+1)*SLG-(X2+3)*OBP] / [(X1+2)*OBP-(X1+1)*SLG]
Y3 = [(1-OBP)*SLG] / [(X1+2)*OBP-(X1+1)*SLG]
D = Y1*HR + Y2*T + Y3*BB
Rearranging the derived S equation for T gives
X2*T = S - X1*D - X3*HR
T = (1/X2)*S - (X1/X2)*D - (X3/X2)*HR
Substituting into this the derived D equation:
T = (1/X2)*S - (X1/X2)*(Y1*HR + Y2*T + Y3*BB) - (X3/X2)*HR
[(1 + (X1*Y2/X2)]*T = (1/X2)*S - (X1*Y3/X2)*BB - [(X3/X2) + (X1*Y1/X2)]*HR
T = [1/(X2+X1*Y2)]*S - [(X1*Y3)/(X2+X2*Y2)]*BB - [(X3+X1*Y1)/(X2+X1*Y2)]*HR
Substituting this into the derived D equation gives
D = Y1*HR + Y3*BB + Y2*{[1/(X2+X1*Y2)]*S - [(X1*Y3)/(X2+X2*Y2)]*BB - [(X3+X1*Y1)/(X2+X1*Y2)]*HR}
D = [Y3 - (X1*Y2*Y3)/(X2+X1*Y2)]*BB + [Y2/(X2+X1*Y2)]*S + [Y1 - Y2*(X3+X1*Y1)/(X2+X1*Y2)]*HR
Y1, Y2, and Y3 can be simplified.
Y1 = [(X3+1)*SLG-(X3+4)*OBP] / [(X1+2)*OBP-(X1+1)*SLG]
Y2 = [(X2+1)*SLG-(X2+3)*OBP] / [(X1+2)*OBP-(X1+1)*SLG]
Y3 = [(1-OBP)*SLG] / [(X1+2)*OBP-(X1+1)*SLG]
Y1 = -3
Y2 = -2
Y3 = (1-OBP) * [(SLG-BA)/(OBP-BA)]
Equation Summary:
X1 = (2*BA-SLG) / (SLG-BA)
X2 = (3*BA-SLG) / (SLG-BA)
X3 = (4*BA-SLG) / (SLG-BA)
Y1 = -3
Y2 = -2
Y3 = (1-OBP) * [(SLG-BA)/(OBP-BA)]
BB = (OBP-BA)/(1-OBP) * AB
S = (0.03298 + 1.341779*AVG - 0.24882*OBP - 0.33932*SLG) * (AB+BB)
D = [Y3 - (X1*Y2*Y3)/(X2+X1*Y2)]*BB + [Y2/(X2+X1*Y2)]*S + [Y1 - Y2*(X3+X1*Y1)/(X2+X1*Y2)]*HR
T = [1/(X2+X1*Y2)]*S - [(X1*Y3)/(X2+X2*Y2)]*BB - [(X3+X1*Y1)/(X2+X1*Y2)]*HR
HR = (0.011458 - 0.35722*AVG - 0.02256*OBP + 0.281748*SLG) * (AB+BB)
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