• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Independent Increments-Ch4

Actuary@22

Very Active Member
Hi
I didn't get the first line on Pg 9 of Ch-4.If the processes are independent so how can we generalise and conclude that the increments are independent as well?
Kindly explain.
Thank you
 
Hello

If process A is independent of process B then changes in process A are independent of changes in process B. ie if the values of process A have no impact on / relationship with process B and vice versa, then the changes in value of process A have no impact on / relationship with the changes of value for process B.

Say we have discrete valued processes then:

\( P(A_{s_1} - A_{t_1} = a, B_{s_2} - B_{t_2} = b) = \Sigma_{v_1} \Sigma_{v_2} P(A_{t_1} = v_1, A_{s_1} = v_1 + a, B_{t_2} = v_2, B_{s_2} = v_2 + b) \)

\( = \Sigma_{v_1} \Sigma_{v_2} P(A_{t_1} = v_1, A_{s_1} = v_1 + a) * P(B_{t_2} = v_2, B_{s_2} = v_2 + b) \)

by independence of \( \{A_t\} \) and \( \{B_t\} \).

\( = \Sigma_{v_1} P(A_{t_1} = v_1, A_{s_1} = v_1 + a) * \Sigma_{v_2} P(B_{t_2} = v_2, B_{s_2} = v_2 + b) \)

\( = P(A_{s_1} - A_{t_1} = a) * P(B_{s_2} - B_{t_2} = b) \)

therefore they are independent.

Hope this helps

Andy
 
Back
Top