# Variances of insurances

98 views
This is a question with respect to the variances of insurances.

Letting:
$$X$$ = PV of term insurance
$$Y$$ = PV of pure endowment
$$Z = X + Y$$ = PV of endowment.

Now, the book says that $$\mathbb{E}[Z^2] = \mathbb{E}[X^2] + \mathbb{E}[Y^2]$$ (Is this correct?)

If this is true it means that $$X$$ and $$Y$$ are independent. It makes sense intuitively in my mind, but could someone please confirm that the above is correct and explain the reasoning behind it?

Reference: Page 92 of the Prescribed Textbook, just under (4.18)

answered Aug 14, 2016 by (1,350 points)
selected Aug 14, 2016

Just to put everything on this page, you have referred to the interpretation of the following from page 92 of the textbook:
$$^2A_{x:\overline{n}|} = \sum_{k=0}^{n-1} v^{2(k+1)} {}_kp_x q_{x+k} + v^{2n} {}_n p_x$$.

So $$X$$  is defined as $$X = v^{K_{x}+1}$$ when $$K_x < n$$ and 0 otherwise.

Also, $$Y$$  is defined as $$Y = v^n$$ when $$K_x \geqslant n$$ and 0 otherwise.

Define $$Z =X +Y$$, as you mentioned.

You are right in saying that the above equation implies that $$\mathbb{E}(Z^2) = \mathbb{E}(X^2) + \mathbb{E}(Y^2)$$, but this does not imply independence of $$X$$ and $$Y$$.

Expanding $$\mathbb{E}(Z^2)$$:

$$\mathbb{E}(Z^2) = \mathbb{E}((X+Y)^2) \Rightarrow \mathbb{E}(Z^2) = \mathbb{E}(X^2) + \mathbb{E}(Y^2) + 2 \mathbb{E}(XY)$$.

The equation given in the textbook only implies that $$\mathbb{E}(XY) = 0$$, so:

$$Cov(X,Y) = \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y) = - \mathbb{E}(X)\mathbb{E}(Y)$$.

So $$X$$ and $$Y$$ are in fact negatively correlated, which makes sense intuitively as they are the payouts payable on death and survival of the life respectively. If $$X$$  is positive, $$Y = 0$$  and vice versa.

+1 vote
answered Aug 14, 2016 by (480 points)

That it is correct. If you imagine what X and Y look like in piece-wise form, then we you multiply X and Y together to evaluate 2E[XY], the product will be zero.