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Finding the value of H when it is given a survival function.

0

$begingroup$

Given $$S(x) = 1 - fracx120, x in [0, 120].$$

If Thomas dies (now he is 30 years old), his family will get $1000000$ with probability $0.2$. At the beginning of each year Thomas pays premium H with probability $0.8$. $v = frac 1 1.06.$

Next in the formulas: $ _kp_x = frac S(k+30) S(30).$

Determine the premium intensity $H$ by adopting the equivalence principle.

My solution:

$H = frac 0.2 times 1000000 times A_30 0.8 times ddot a_30$.

The problem is that I cannot calculate $ddot a_30$ or $A_30$, because the formulas are:

$ddot a_30 = sum_k=0^∞ v^k times _kp_30$ or

$A_30 = sum_k=0^∞ v^k+1 times _kp_30 times q_30+k.$

So here I have an infinity. I need to find one of these values at least, because I can find the second value using formula $ddot a_30 = frac 1 - A_30 1-v$. But I do not know how to do that (probably the $S(x)$ need to be used). Maybe I do not know some formulas.

The answer have to be $H=3195.$

probability

share|cite|improve this question

edited Mar 27 at 16:51

Philip

asked Mar 27 at 16:10

PhilipPhilip

917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not able to relate $S(x)$ with the sum his family gets or the probability with which they will. What am I missing here? Nor do I understand the terms $A_30$ and $ddota_30$ in your solution. Can you add some prose to describe them?
  $endgroup$
  – Balakrishnan Rajan
  Mar 27 at 16:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  S(x) can be related only with $_kp_30 = frac s(30+k) S(30).$
  $endgroup$
  – Philip
  Mar 27 at 16:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK..... so what's that?
  $endgroup$
  – Balakrishnan Rajan
  Mar 27 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fraction for finding H is correct, my professor started solving this exercise.
  $endgroup$
  – Philip
  Mar 27 at 16:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have a problem only with calculating that sum from 0 to infinity, I am not getting the value.
  $endgroup$
  – Philip
  Mar 27 at 16:33
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

Given $$S(x) = 1 - fracx120, x in [0, 120].$$

If Thomas dies (now he is 30 years old), his family will get $1000000$ with probability $0.2$. At the beginning of each year Thomas pays premium H with probability $0.8$. $v = frac 1 1.06.$

Next in the formulas: $ _kp_x = frac S(k+30) S(30).$

Determine the premium intensity $H$ by adopting the equivalence principle.

My solution:

$H = frac 0.2 times 1000000 times A_30 0.8 times ddot a_30$.

The problem is that I cannot calculate $ddot a_30$ or $A_30$, because the formulas are:

$ddot a_30 = sum_k=0^∞ v^k times _kp_30$ or

$A_30 = sum_k=0^∞ v^k+1 times _kp_30 times q_30+k.$

So here I have an infinity. I need to find one of these values at least, because I can find the second value using formula $ddot a_30 = frac 1 - A_30 1-v$. But I do not know how to do that (probably the $S(x)$ need to be used). Maybe I do not know some formulas.

The answer have to be $H=3195.$

probability

share|cite|improve this question

edited Mar 27 at 16:51

Philip

asked Mar 27 at 16:10

PhilipPhilip

917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not able to relate $S(x)$ with the sum his family gets or the probability with which they will. What am I missing here? Nor do I understand the terms $A_30$ and $ddota_30$ in your solution. Can you add some prose to describe them?
  $endgroup$
  – Balakrishnan Rajan
  Mar 27 at 16:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  S(x) can be related only with $_kp_30 = frac s(30+k) S(30).$
  $endgroup$
  – Philip
  Mar 27 at 16:29
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK..... so what's that?
  $endgroup$
  – Balakrishnan Rajan
  Mar 27 at 16:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The fraction for finding H is correct, my professor started solving this exercise.
  $endgroup$
  – Philip
  Mar 27 at 16:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have a problem only with calculating that sum from 0 to infinity, I am not getting the value.
  $endgroup$
  – Philip
  Mar 27 at 16:33
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

Given $$S(x) = 1 - fracx120, x in [0, 120].$$

If Thomas dies (now he is 30 years old), his family will get $1000000$ with probability $0.2$. At the beginning of each year Thomas pays premium H with probability $0.8$. $v = frac 1 1.06.$

Next in the formulas: $ _kp_x = frac S(k+30) S(30).$

Determine the premium intensity $H$ by adopting the equivalence principle.

My solution:

$H = frac 0.2 times 1000000 times A_30 0.8 times ddot a_30$.

The problem is that I cannot calculate $ddot a_30$ or $A_30$, because the formulas are:

$ddot a_30 = sum_k=0^∞ v^k times _kp_30$ or

$A_30 = sum_k=0^∞ v^k+1 times _kp_30 times q_30+k.$

So here I have an infinity. I need to find one of these values at least, because I can find the second value using formula $ddot a_30 = frac 1 - A_30 1-v$. But I do not know how to do that (probably the $S(x)$ need to be used). Maybe I do not know some formulas.

The answer have to be $H=3195.$

probability

share|cite|improve this question

edited Mar 27 at 16:51

Philip

asked Mar 27 at 16:10

PhilipPhilip

917

$endgroup$

share|cite|improve this question

edited Mar 27 at 16:51

Philip

asked Mar 27 at 16:10

PhilipPhilip

917

share|cite|improve this question

edited Mar 27 at 16:51

Philip

asked Mar 27 at 16:10

PhilipPhilip

917

asked Mar 27 at 16:10

PhilipPhilip

917

asked Mar 27 at 16:10

PhilipPhilip

917

1 Answer
1

active

oldest

votes

0

$begingroup$

Given that,
$$
ddota_30 = sum_k=0^inftyleft(frac11.06right)^k times fracS(30+k)S(30)
$$

Now as $S(30) = 0.75$, the sum is,
$$
ddota_30 = frac10.75left( sum_k=0^infty left(frac11.06right)^k left( 1 - frac30+k120right) right)=left( sum_k=0^infty left(frac11.06right)^k - frac190 sum_k=0^infty kleft(frac11.06right)^k right)
$$

The first sum is a sum to $infty$ of a GP. The second sum can be obtained from How to derive the closed form of the sum of $kr^k$
?

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

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0

$begingroup$

Given that,
$$
ddota_30 = sum_k=0^inftyleft(frac11.06right)^k times fracS(30+k)S(30)
$$

Now as $S(30) = 0.75$, the sum is,
$$
ddota_30 = frac10.75left( sum_k=0^infty left(frac11.06right)^k left( 1 - frac30+k120right) right)=left( sum_k=0^infty left(frac11.06right)^k - frac190 sum_k=0^infty kleft(frac11.06right)^k right)
$$

The first sum is a sum to $infty$ of a GP. The second sum can be obtained from How to derive the closed form of the sum of $kr^k$
?

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

0

$begingroup$

Given that,
$$
ddota_30 = sum_k=0^inftyleft(frac11.06right)^k times fracS(30+k)S(30)
$$

Now as $S(30) = 0.75$, the sum is,
$$
ddota_30 = frac10.75left( sum_k=0^infty left(frac11.06right)^k left( 1 - frac30+k120right) right)=left( sum_k=0^infty left(frac11.06right)^k - frac190 sum_k=0^infty kleft(frac11.06right)^k right)
$$

The first sum is a sum to $infty$ of a GP. The second sum can be obtained from How to derive the closed form of the sum of $kr^k$
?

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

add a comment | 

$begingroup$

Given that,
$$
ddota_30 = sum_k=0^inftyleft(frac11.06right)^k times fracS(30+k)S(30)
$$

Now as $S(30) = 0.75$, the sum is,
$$
ddota_30 = frac10.75left( sum_k=0^infty left(frac11.06right)^k left( 1 - frac30+k120right) right)=left( sum_k=0^infty left(frac11.06right)^k - frac190 sum_k=0^infty kleft(frac11.06right)^k right)
$$

The first sum is a sum to $infty$ of a GP. The second sum can be obtained from How to derive the closed form of the sum of $kr^k$
?

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

$endgroup$

share|cite|improve this answer

edited Mar 27 at 17:50

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210

answered Mar 27 at 17:07

Balakrishnan RajanBalakrishnan Rajan

15210