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Calculation example for a population founded on a single philogenetic tree

At the first generation there is the group G0={e1,e2,e3} with e2 having the same genotype as e3. The maximal number of child is 2. The distribution for e1 in all groups is given with d0=1/4 d1=3/4 d2=0. The distribution for e2 and e3 is for all groups but {e1,e2} 1/2 1/3 1/6. For the group {e1,e2} it is given with 1/7 4/7 2/7.

What is the probability of the population {e2} in the second generation?

Because e3 and e2 have the same genotype e2 is equated with e3.

This 15 groups that can come frome G0:
{},{e2},{e2,e2},{e2,e2,e2},{e2,e2,e2,e2},
{e1},{e1,e2},{e1,e2,e2},{e1,e2,e2,e2},{e1,e2,e2,e2,e2}
{e1,e1},{e1,e1,e2},{e1,e1,e2,e2},{e1,e1,e2,e2,e2},{e1,e1,e2,e2,e2,e2}

The probability of the phylogenies                             (1/4 3/4 0 , 1/2 1/3 1/6 , 1/7 4/7 2/7)
{e1,e2,e2}->{}->{e2} ..*0 0
{e1,e2,e2}->{e2}->{e2} ( (1/4*1/2*1/3)*2)*1/3 1/36
{e1,e2,e2}->{e2,e2}->{e2} ( (1/4*1/2*1/6)*2+(1/4*1/3*1/3))*1/2*1/3*2 5/216
{e1,e2,e2}->{e2,e2,e2}->{e2} ( (1/4*1/3*1/6)*2)*1/2*1/2*1/3*3 1/144
{e1,e2,e2}->{e2,e2,e2,e2}->{e2} (1/4*1/6*1/6)*1/2*1/2*1/2*1/3*4 1/864
{e1,e2,e2}->{e1}->{e2} ..*0 0
{e1,e2,e2}->{e1,e2}->{e2} ( (3/4*1/2*1/3)*2)*1/4*4/7 1/28
{e1,e2,e2}->{e1,e2,e2}->{e2} ( (3/4*1/2*1/6)*2+(3/4*1/3*1/3))*1/4*1/2*1/3*2 5/288
{e1,e2,e2}->{e1,e2,e2,e2}->{e2} ( (3/4*1/3*1/6)*2)*1/4*1/2*1/2*1/3*3 1/192
{e1,e2,e2}->{e1,e2,e2,e2,e2}->{e2} ( (3/4*1/6*1/6)*1/4*1/2*1/2*1/2*1/3*4 1/1152
{e1,e2,e2}->{e1,e1}->{e2} 0*0 0
{e1,e2,e2}->{e1,e1,e2}->{e2} 0*.. 0
{e1,e2,e2}->{e1,e1,e2,e2}->{e2} 0*.. 0
{e1,e2,e2}->{e1,e1,e2,e2,e2}->{e2} 0*.. 0
{e1,e2,e2}->{e1,e1,e2,e2,e2,e2}->{e2} 0*.. 0


By using wkat1kbt2 = ( (d*..+p+..)*j*..)*t*..+k+..      (1/4 3/4 0 , 1/2 1/3 1/6 , 1/7 4/7 2/7)
((1/4)*(1/2*1/2))*((1)*(0)) 0
+((1/4)*(1/2*1/3+1/3*1/2))*((1)*(1/3)) 1/36
+((1/4)*(1/2*1/6+1/3*1/3+1/6*1/2))*((1)*(1/2*1/3+1/3*1/2)) 5/216
+((1/4)*(1/3*1/6+1/6*1/3))*((1)*(1/2*1/2*1/3+1/2*1/3*1/2+1/3*1/2*1/2))
1/144
+((1/4)*(1/6*1/6))*((1)*(1/2*1/2*1/2*1/3+1/2*1/2*1/3*1/2+1/2*1/3*1/2*1/2+1/3*1/2*1/2*1/2))
1/864
+((3/4)*(1/2*1/2))*((1/4)*(0)) 0
+((3/4)*(1/2*1/3+1/3*1/2))*((1/4)*(4/7)) 1/28
+((3/4)*(1/2*1/6+1/3*1/3+1/6*1/2))*((1/4)*(1/2*1/3+1/3*1/2)) 5/288
+((3/4)*(1/3*1/6+1/6*1/3))*((1/4)*(1/2*1/2*1/3+1/2*1/3*1/2+1/3*1/2*1/2)) 1/192
+((3/4)*(1/6*1/6))*((1/4)*(1/2*1/2*1/2*1/3+1/2*1/2*1/3*1/2+1/2*1/3*1/2*1/2+1/3*1/2*1/2*1/2)) 1/1152
+((0)*(1/2*1/2))*((1/4*1/4)*(0)) 0
+((0)*(1/2*1/3+1/3*1/2))*((1/4*1/4)*(1/3)) 0
+((0)*(1/2*1/6+1/3*1/3+1/6*1/3))*((1/4*1/4)*(1/2*1/3+1/3*1/2)) 0
+((0)*(1/3*1/6+1/6*1/3))*((1/4*1/4)*(1/2*1/2*1/3+1/2*1/3*1/2+1/3*1/2*1/2)) 0
+((0)*(1/6*1/6))*((1/4*1/4)*(1/2*1/2*1/2*1/3+1/2*1/2*1/3*1/2+1/2*1/3*1/2*1/2+1/3*1/2*1/2*1/2)) 0

wkat1kbt2 = S[k0,k1,..,kt2-t1 : k0=ka; k2..kt2-t1-1 >= 0; kt2-t1=kb]P[t : 0<=t<t2-t1]P[j]vjktkt+1t1+t
=S[k0,k1,..,kt2-t1 : k0=ka; k2..kt2-t1-1 >= 0; kt2-t1=kb]P[t : 0<=t<t2-t1]P[j]S[p1,..,px : x=N(e'jkt)] 1(y=N(e'jkt+1),p1,..,px)*dp1ijktt*..*dpxijktt

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