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for question 9.8 in DHW: how do we know or prove that for independent lives that \(u_{x+t:y+t} = u_{x+t} + u_{y+t}\)

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asked Nov 15, 2017 in BUS 3024S - Contingencies by adam (180 points)

The question ask to show that the density of \(T_{xy}\) is \( _tp_{xy}(u_{x+t}+u_{y+t})\)

1 Answer

+1 vote
answered Nov 21, 2017 by Feroz (1,390 points)

First, we find the distribution function of \(T_{xy}\):

$$Pr(T_{xy} \leq t) = 1 - Pr(T_{xy} > t) = 1 - Pr(T_{x} > t) \cdot Pr(T_{y} > t) = 1 - {}_t p_{x} {}_t p_{y}$$.

To find the density, we differentiate the above expression with respect to \(t\). 

$$f_{T_{xy}}(t) = \frac{d}{dt} Pr(T_{xy} \leq t) = {}_t p_{xy}(\mu_{x+t}+ \mu_{y+t})$$ (you need to use the product rule to differentiate the second term in the expression for \(Pr(T_{xy} \leq t) \)).

You will also need expressions of the following form:

$$\frac{d}{dt} {}_t p_{x} = -{}_t p_{x} \mu_{x+t}$$.

Hope this helps.

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