Let $$Λ_1 = \frac{1}{4} Λ^{(2)} + \frac{3}{4} Λ^{(4)},$$
$Λ^{(2)}, Λ^{(4)}$ are independent random variables.
$\mathbb EΛ^{(s)}=\frac{s+5}{s^2}$ and $\mathbb DΛ^{(s)}=\frac{(s+5)^2}{s^2}.$
From the risk theory $\mathbb D(L_1') = \mathbb E Λ_1 + \mathbb D Λ_1$ or $\mathbb D(L_1') = Cov(Λ_1,Λ_1).$
Find $\mathbb D(L_1').$
Solution:
I tried to solve this in 2 different ways (used 2 formulas mentioned before).
1) $\mathbb D(L_1') = \mathbb E Λ_1 + \mathbb D Λ_1 = \frac{55}{64} + \mathbb D(\frac{1}{4} Λ^{(2)} + \frac{3}{4} Λ^{(4)}) = \frac{55}{64} + \frac{1}{16}\mathbb D Λ^{(2)} + \frac{9}{16} \mathbb DΛ^{(4)} = 4\frac{121}{256}. $
2) $\mathbb D(L_1') = Cov(Λ_1,Λ_1)= Cov(\frac{1}{4} Λ^{(2)} + \frac{3}{4} Λ^{(4)},\frac{1}{4} Λ^{(2)} + \frac{3}{4} Λ^{(4)}) = 3 \frac{157}{256},$
here $Cov(Λ^{(i)},Λ^{(j)})=0,$ if $i\neq j.$
But I get different answers. What is wrong?