Complete Expectation of Life assuming UDD

[willborg35]

Under UDD it is assumed exactly that:

Complete expectation of Life = Kurate Expectation of Life + 1/2

Does this apply to the limited expectation of life (related complte(ex:n) =ex:n + 1/2)?

Does anyone know a formula for the limited life expectancy formulas under UDD? Thank you.

[MathForMarines]

I don't recall ever having seen one, but I have also done every life con question I could get my hands on, and I never was in a situation in which I would need a formula like that.


Under UDD it is assumed exactly that:

Complete expectation of Life = Kurate Expectation of Life + 1/2

Does this apply to the limited expectation of life (related complte(ex:n) =ex:n + 1/2)?

Does anyone know a formula for the limited life expectancy formulas under UDD? Thank you.

[ctperng]

Under UDD it is assumed exactly that:

Complete expectation of Life = Kurate Expectation of Life + 1/2

Does this apply to the limited expectation of life (related complte(ex:n) =ex:n + 1/2)?
Does anyone know a formula for the limited life expectancy formulas under UDD? Thank you.

There is an approximation: you add (1/2)*(n_q_x) instead of 1/2.

ctperng

[NoMoreExams]

Under UDD it is assumed exactly that:

Complete expectation of Life = Kurate Expectation of Life + 1/2

Does this apply to the limited expectation of life (related complte(ex:n) =ex:n + 1/2)?

Does anyone know a formula for the limited life expectancy formulas under UDD? Thank you.

This is easy to figure out

Complete: e_{x:n} = integral(t_p_x dt from 0 to n) = integral(1 - t/(w-x) dt from 0 to n) = n - n^2/(2(w-x)) = n[1 - n/[2(w-x)]]

Curtate (not Kurate): e_{x:n} = sum(t_p_x from t=1 to n - 1) = sum(1 - k/(w-x) from 1 to n-1) = (n-1) - n(n-1)/[2(w-x)] = (n-1)[1 - n/[2(w-x)]] = n[1 - n/[2(w-x)]] - [1 - n/[2(w-x)]]

So as you can see Curtate = Complete - [1 - n/[2(w-x)]]

Obviously if the sum goes to "infinity" i.e. in this case to w - x, we let n = w-x and we get:

Curtate = Complete- [1 - (w-x)/[2(w-x)]] = Complete - [1 - 1/2] = Complete - 1/2

Hopefully my algebra is right.
________
jugallette (http://juggalos.org)

[zmkramer]

This is easy to figure out

Complete: e_{x:n} = integral(t_p_x dt from 0 to n) = integral(1 - t/(w-x) dt from 0 to n) = n - n^2/(2(w-x)) = n[1 - n/[2(w-x)]]

Curtate (not Kurate): e_{x:n} = sum(t_p_x from t=1 to n - 1) = sum(1 - k/(w-x) from 1 to n-1) = (n-1) - n(n-1)/[2(w-x)] = (n-1)[1 - n/[2(w-x)]] = n[1 - n/[2(w-x)]] - [1 - n/[2(w-x)]]

So as you can see Curtate = Complete - [1 - n/[2(w-x)]]

Obviously if the sum goes to "infinity" i.e. in this case to w - x, we let n = w-x and we get:

Curtate = Complete- [1 - (w-x)/[2(w-x)]] = Complete - [1 - 1/2] = Complete - 1/2

Hopefully my algebra is right.

I got something else, when you do the n-year curtate temporary life expectancy you get

summation of (tPx)

which is 1 - k/(w-x) from 1 to n (NOT n-1)

solving the summation would give you

n - (summation of k from 1 to n)/(w-x) this would give you

n - n(n+1)/2(w-x) which is n - n^2/2(w-x) - n/2(w-x)

so that is continuous e_x:n = curtate e_x:n + n/2(w-x) which would only be 1/2 when n = w-x


which is what ctperng was saying

[NoMoreExams]

I got something else, when you do the n-year curtate temporary life expectancy you get

summation of (tPx)

which is 1 - k/(w-x) from 1 to n (NOT n-1)

solving the summation would give you

n - (summation of k from 1 to n)/(w-x) this would give you

n - n(n+1)/2(w-x) which is n - n^2/2(w-x) - n/2(w-x)

so that is continuous e_x:n = curtate e_x:n + n/2(w-x) which would only be 1/2 when n = w-x


which is what ctperng was saying

booyah

I agree, my summation should go to n not n - 1, however our results agree, when you plug in n = w - x, you get cont = curtate + 1/2, I wrote down curtate = cont - 1/2.

[R_T]

Under UDD it is assumed exactly that:

Complete expectation of Life = Kurate Expectation of Life + 1/2

Does this apply to the limited expectation of life (related complte(ex:n) =ex:n + 1/2)?

Does anyone know a formula for the limited life expectancy formulas under UDD? Thank you.

Actually, it's ex:n + .5(1-nPx).