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Thomas H
March 3rd 2007, 06:49 PM
What is the summation form of the accumulated due value s_{x:n}?

There is a formula given in the study manual in lesson 17 that looks like:

accumulated-due s_{x:n} = annuity-due a_{x:n}/ n_E_x

I keep trying to understand this by looking at the summation form of annuity-due a_{x:n}.

annuity-due a_{x:n} = sum(k=1 to n) v^(k-1) * {k-1}_p_x

where v = 1/(1+i). I divide this by n_E_x = v^n * n_p_x but my result doesn't seem to make sense to me.

wat
March 5th 2007, 02:46 PM
What is the summation form of the accumulated due value s_{x:n}?

There is a formula given in the study manual in lesson 17 that looks like:

accumulated-due s_{x:n} = annuity-due a_{x:n}/ n_E_x

I keep trying to understand this by looking at the summation form of annuity-due a_{x:n}.

annuity-due a_{x:n} = sum(k=1 to n) v^(k-1) * {k-1}_p_x

where v = 1/(1+i). I divide this by n_E_x = v^n * n_p_x but my result doesn't seem to make sense to me.

There's no easy way to interpret the probability portion of the summation of the accumulated-due.

If you divide each term by v^n * n_p_x, however, you can simplify it a little, so that the summation becomes:

s_{x:n} = (1 / n_p_x) sum[k=1 to n] (1+i)^(n-(k-1)) * {k-1}_p_x.

Thomas H
March 5th 2007, 04:25 PM
There's no easy way to interpret the probability portion of the summation of the accumulated-due.

Thanks. This explains my problem of understanding the summation form I was ending up with.

wat
March 5th 2007, 04:28 PM
Thanks. This explains my problem of understanding the summation form I was ending up with.

I don't mean to be vague about it, but in short, if you divide each probability ( {k-1}_p_x ) by n_p_x, then each factor will be > 1, which implies that the factor is no longer a probability.