Two equalities in insurance mathematics I'm using classical actuarial notation. $(IA)_x$ is APV of increasing insurance for $x$-years old person that pays $k$ when the person dies in year $k$. Also $(DA)_{1_{x:\bar{n|}}}$ is APV of decresing insurence for $x$-years old person that pays $n-k+1$ in case when person dies in year $k$. I need to prove these two things:

*

*$(IA)_x=\nu q_x+\nu p_x(A_{x+1}+(IA)_{x+1});$

*$(DA)_{1_{x:\bar{n|}}}=(n+1)A_{1_{x:\bar{n|}}}-(IA)_{1_{x:\bar{n|}}}$
I think I have the first one solved:
$(IA)_x=\sum_{k=0}^\infty (k+1)\nu^{k+1}\cdot {}_k p_x \cdot q_{x+k}$
$A_{x+1}=\sum_{k=0}^\infty \nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}$
So on the right side we have:
$\nu q_x+\nu p_x(\sum_{k=0}^\infty [\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}]+\sum_{k=0}^\infty [(k+1)\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k}])=$
$=\nu q_x+\nu p_x(\sum_{k=0}^\infty(k+2)\nu^{k+1}\cdot {}_k p_{x+1} \cdot q_{x+1+k})=\nu q_x+(\sum_{k=0}^\infty(k+2)\nu^{k+2}\cdot {}_{k+1} p_{x} \cdot q_{x+1+k})=$
$=\nu q_x+(\sum_{k=0}^\infty(k+1)\nu^{k+1}\cdot {}_{k} p_{x} \cdot q_{x+k})=\sum_{k=0}^\infty (k+1)\nu^{k+1}\cdot {}_k p_x \cdot q_{x+k}=(IA)_x.$
Is that correct? Could anyone give me a hint how to prove the second equality?
 A: The intuitive way to understand $$(IA)_x = v q_x + vp_x (A_{x+1} + (IA)_{x+1}) \tag{1}$$ is to condition on the survival of $(x)$ after one year.  Either $(x)$ dies in the first year with probability $q_x$, in which case the insurance pays $1$ at the end of the first year, which has present value $v$, or $(x)$ survives one year with probability $p_x$.  In this second case, we express the insurance payment as the sum of a whole life insurance of $1$, plus an increasing insurance, on $(x+1)$.  Then we discount by $v$ because one year has already passed.
Consequently, we write the RHS of Equation $(1)$ as
$$\begin{align}
vq_x + vp_x(A_{x+1} + (IA)_{x+1})
&= vq_x + vp_x \sum_{k=0}^\infty \left( 1 + (k+1) \right) v^{k+1} ({}_k p_{x+1})( q_{x+k+1}) \\
&= vq_x + \sum_{k=0}^\infty (k+2) v^{k+2} (p_x) ({}_k p_{x+1}) (q_{x+k+1}) \\
&= vq_x + \sum_{k=0}^\infty (k+2) v^{k+2} ({}_{k+1} p_x) (q_{x+k+1}) \\
&= vq_x + \sum_{k=1}^\infty (k+1) v^{k+1} ({}_k p_x) (q_{x+k} ) \\
&= \sum_{k=0}^\infty (k+1) v^{k+1} ({}_k p_x) (q_{x+k}) \\
&= (IA)_x.
\end{align}$$
This is essentially the same as your proof.
For the identity
$$\require{enclose}
(DA)_{x:\enclose{actuarial}{n}}^1 = (n+1) A_{x:\enclose{actuarial}{n}}^1 - (IA)_{x:\enclose{actuarial}{n}}^1, \tag{2}$$ it is more convenient to move the increasing insurance on the RHS to the left:
$$(DA)_{x:\enclose{actuarial}{n}}^1 + (IA)_{x:\enclose{actuarial}{n}}^1 = (n+1) A_{x:\enclose{actuarial}{n}}^1 \tag{3}$$
and now it becomes obvious what this equation means:  the sum of a $n$-year decreasing insurance on $(x)$ plus a $n$-year increasing insurance on $(x)$ equals a $n$-year insurance on $(x)$ that pays $n+1$ at any time during the term of the policy.  Think of the schedule of payments on the LHS:  given that $(x)$ dies in year $k$, for any $k \in \{1, 2, \ldots, n\}$, the decreasing insurance pays $n-k+1$, but the increasing insurance pays $k$, so the present value of the LHS is $(n+1)v^k$, which is also the present value of the RHS.  This should provide you with sufficient motivation to formulate a proof using summation notation.
