Prove that deferred continuous annuity has APV: $E(e^{-int_w^t delta(s)ds}int_0^{T-w}e^{-int_w^{w+t}...
I want to prove that deferred continuous annuity has the APV:
$$E(e^{-int_w^t delta(s)ds}int_0^{T-w}e^{-int_w^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
$$=E(int_0^{T-w}e^{-int_0^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
Where $T$ is the lifetime RV, $w$ is the point at which the annuity starts $delta$ is the interest.
The formula is quite obvious, but how do I prove it? What I need to show is that it equals:
$$text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)+mu(x+m)ds}$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }P(T>w+k/m)$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }E(mathbb I_{T>w+k/m})$$
Or at least equals:
$$int_w^{infty}e^{-int_0^{t} delta(s)ds}mathbb P(T>t)dt$$
The issue is, how do I take the expectation with regards to the integral?
actuarial-science
asked Nov 29 '18 at 12:10
Dole
904514
add a comment |
I want to prove that deferred continuous annuity has the APV:
$$E(e^{-int_w^t delta(s)ds}int_0^{T-w}e^{-int_w^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
$$=E(int_0^{T-w}e^{-int_0^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
Where $T$ is the lifetime RV, $w$ is the point at which the annuity starts $delta$ is the interest.
The formula is quite obvious, but how do I prove it? What I need to show is that it equals:
$$text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)+mu(x+m)ds}$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }P(T>w+k/m)$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }E(mathbb I_{T>w+k/m})$$
Or at least equals:
$$int_w^{infty}e^{-int_0^{t} delta(s)ds}mathbb P(T>t)dt$$
The issue is, how do I take the expectation with regards to the integral?
actuarial-science
asked Nov 29 '18 at 12:10
Dole
904514
add a comment |
I want to prove that deferred continuous annuity has the APV:
$$E(e^{-int_w^t delta(s)ds}int_0^{T-w}e^{-int_w^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
$$=E(int_0^{T-w}e^{-int_0^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
Where $T$ is the lifetime RV, $w$ is the point at which the annuity starts $delta$ is the interest.
The formula is quite obvious, but how do I prove it? What I need to show is that it equals:
$$text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)+mu(x+m)ds}$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }P(T>w+k/m)$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }E(mathbb I_{T>w+k/m})$$
Or at least equals:
$$int_w^{infty}e^{-int_0^{t} delta(s)ds}mathbb P(T>t)dt$$
The issue is, how do I take the expectation with regards to the integral?
actuarial-science
asked Nov 29 '18 at 12:10
Dole
904514
I want to prove that deferred continuous annuity has the APV:
$$E(e^{-int_w^t delta(s)ds}int_0^{T-w}e^{-int_w^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
$$=E(int_0^{T-w}e^{-int_0^{w+t} delta(s)ds}dtmathbb I_{T>w})$$
Where $T$ is the lifetime RV, $w$ is the point at which the annuity starts $delta$ is the interest.
The formula is quite obvious, but how do I prove it? What I need to show is that it equals:
$$text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)+mu(x+m)ds}$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }P(T>w+k/m)$$
$$=text{lim}_{mrightarrow infty }frac{1}{m}sum_{k=0}^infty e^{-int_0^{w+k/m} delta(s)ds }E(mathbb I_{T>w+k/m})$$
Or at least equals:
$$int_w^{infty}e^{-int_0^{t} delta(s)ds}mathbb P(T>t)dt$$
The issue is, how do I take the expectation with regards to the integral?
actuarial-science
actuarial-science
asked Nov 29 '18 at 12:10
Dole
904514
asked Nov 29 '18 at 12:10
Dole
904514
edited Nov 29 '18 at 12:26
asked Nov 29 '18 at 12:10
Dole
904514
asked Nov 29 '18 at 12:10
Dole
904514
asked Nov 29 '18 at 12:10
Dole
904514
904514
add a comment |
add a comment |
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