Let $Y$ be a present value of whole life annuity that pays $1$ for year of each $x$-years old. We know that $\ddot{a}_x=10$ for $i'=1/24=e^\delta-1$ and $\ddot{a}_x=6$ for $i=e^{2\delta}-1$. I want to calculate the variance of $Y$. To be honest I don't know where to start.
I want to use the equation:
$$VarY=\frac{^2A_x-(A_x)^2}{d^2}.$$ Where: $$A_x=\sum_{j=0}^\infty v^{j+1} {}_j| q_{x};$$ $$^2 A_x=\sum_{j=0}^\infty (v^2)^{j+1} {}_j| q_{x}$$ Also we have: $\ddot{a}_x=\frac{1-A_x}{d}$ where $d=\frac{i}{1+i}$
The definition of $\ddot{a}_x=\sum_{j=0}^\infty v^j\ _{j}p_x$ and $v=\frac{1}{1+i}$.
All I can think of is to calculate $\delta$ and other constants as $v$ and $d$ for $i'$ and $i$ but I don't really know where it should go from there.
You can solve for $A_x$ and ${}^2 A_x$ using the formula $\ddot{a}_x = \frac{1-A_x}{d}$ with the two given values for $\ddot{a}_x$. Then you can solve for $\delta$ using the given $e^\delta - 1 = 1/24$ and use that to solve for $d$.