0

For a de Moivre's (i mean, that follows de Moivre's law) population with unknown parameter $\omega$, knowing $\dot{e}_{x+1}$ is given. From the recursive formula find $\dot{e}_x$

I am not certain which formula should i use here. Appreciate any suggestions.

Any ideas? I would appreciate any viable help.

Clarinetist
  • 19,519
mkropkowski
  • 1,131

1 Answers1

1

Actually, even though you've written $\dot{e}_x$, what I assume you've meant to write is $$\mathring{e}_x\text{.}$$ Let $T_x$ be the time-to-death random variable for $(x)$. Then, a very well-known formula is $$\mathring{e}_x = \mathbb{E}[T_x] = \dfrac{\omega - x}{2}\text{.}$$ Then it follows that $$\mathring{e}_{x+1} = \dfrac{\omega-(x+1)}{2} = \dfrac{\omega-x}{2}-\dfrac{1}{2}=\mathring{e}_x-\dfrac{1}{2}\text{.}$$ This really only works well for the uniform distribution. When people generally refer to the "recursive" formula for $\mathring{e}_x$, they mean $$\mathring{e}_x = \mathring{e}_{x:\overline{n}|}+{}_{n}p_{x}\cdot\mathring{e}_{x+n}\text{.}$$ If $\mathring{e}_{x+1}$ is given, $$\mathring{e}_x = \mathring{e}_{x:\overline{1}|}+p_{x} \cdot \mathring{e}_{x+1}\text{.}$$ Now recall that the PDF of $T_x$ is given by $$f_{T}(t) = \dfrac{1}{\omega - x}\text{, } t \in [0, \omega - x]\text{.}$$ Then, another formula you should have memorized by now is $${}_{t}p_{x} = \dfrac{\omega - x - t}{\omega - x}$$ from which you can find $p_x$ easily, and $$\mathring{e}_{x:\overline{1}|} = \int_{0}^{1}{}_{t}p_{x}\text{ d}t\text{.}$$

Clarinetist
  • 19,519
  • One more question it seems that if i have e.g. $\dot{e}_{x+1}=14$ and that is all i know it seems that still i do not have sufficient information as i am left at the end with unknown x and $\omega$. – mkropkowski May 18 '16 at 07:42
  • @mkropkowski I don't understand. Did you not see my proof up there? If you have $\mathring{e}{x+1}$, you can easily calculate $\mathring{e}_x$. You don't have to have $x$ and $\omega$ to calculate $\mathring{e}_x$ if you have $\mathring{e}{x+1}$. – Clarinetist May 18 '16 at 10:35
  • @mkropkowski Especially on actuarial questions, you can't be expected to know everything. – Clarinetist May 18 '16 at 10:36
  • Possibly i dont fully understand it i mean using the equation you have provided i am still left with variables left possibly i am doing something wrong... I am trying to use this equation without minus half – mkropkowski May 18 '16 at 14:39