Solve[{a==((-m)/(2 l)),b==(1/(2 l)),m==((p0 (su^2)-a b S0)/((su^2)+(b^2)S0)),l==b (S0/((su^2)+(b^2)S0))},{a,b,m,l}]

Exercise 7.1

S0 -(((( b S0 + suv ))^2)/(((su^2)+(b^2) S0+2 b suv)))//Together

x=((((v-μ) (Subscript[Σ, 0])-λ (Subscript[σ, uv]) (v-(Subscript[p, 0]))))/(2 λ (Subscript[Σ, 0])));
t1=Coefficient[x,v,0](* this is alpha *)
 t2=Coefficient[x,v,1] (* this is beta *)

Evy=(Subscript[p, 0])+(((β (Subscript[Σ, 0])+(Subscript[σ, uv])))/((((Subscript[σ, u])^2)+(β^2)(Subscript[Σ, 0])+2 β (Subscript[σ, uv]))))(y-α-β (Subscript[p, 0]));
t3=Simplify[Coefficient[Evy,y,0]] (* this is mu *)
t4=Simplify[Coefficient[Evy,y,1]] (* this is lambda *)

wSol=Solve[{α==t1,β==t2,μ==t3,λ==t4},{α,β,λ,μ}]//Simplify

We must have β>0 so we take the first solution.  Lets check we get the old results if we take (Subscript[σ, uv])=0

Simplify[wSol[[1]]/.{(Subscript[σ, uv])->0},{(Subscript[Σ, 0])>0,(Subscript[σ, u])>0}]

What if u and v have perfect correlation between the

Simplify[wSol[[1]]/.{(Subscript[σ, uv])->(Subscript[σ, u]) Sqrt[(Subscript[Σ, 0])]},{(Subscript[Σ, 0])>0,(Subscript[σ, u])>0}]

Exercise 7.2

Simplify[Solve[{a==((((se^2)p0-((se^2)+S0)m))/(2 l ((se^2)+S0))),b==(S0/(2l ((se^2)+S0))),l==b (S0/((su^2)+(b^2)S0)),m==p0-b (S0/((su^2)+(b^2)S0))(a+b p0)},{a,b,l,m}]]

Exercise 7.3

Solve[{a==((-m)/(2 l(1+g))),b==(1/(2 l(1+g))),l==(((1+g)b S0)/((su^2)+(((1+g))^2)(b^2)S0)),m==p0-(((1+g))^2)b S0( (((a+b p0))/((su^2)+(((1+g))^2)(b^2) S0)))},{a,b,l,m}]//Simplify