-1

I have a state which is a list of probabilities for next states as follows:

$\left[0, \frac {1}{2}, 0, 0, \frac{1}{2}\right]$

I realise that the second element here actually has a $\left(\frac 2 7\right)$ chance of looping back to this state. So I can remove it like so:

$\frac 1 2 -\left(\frac 1 2 \times \frac 2 7\right)$

Which gives me $\frac 5 {14}$, which is correct in this case, but I might still have the wrong method for others

I want the other element to equal $\frac 9 {14}$, which I have been able to do, but not in the general case.

Hope this is clearer.

Hanul Jeon
  • 27,376
DaraJ
  • 113
  • Ummm: $P_2 = 1 - P_1$ perhaps? – David G. Stork Dec 05 '20 at 20:13
  • No that doesn't work. Maybe I didn't explain the question clearly enough. I will edit – DaraJ Dec 05 '20 at 20:14
  • 2
    This single example does not provide enough information about what you "want" for an answer in general, and why you want it. Please [edit] the question to show us more examples (maybe some with $n >2$) and the context in which this want arises. How does this differ from your previous question, where you accepted an answer: https://math.stackexchange.com/questions/3933916/normalise-rest-of-list ? – Ethan Bolker Dec 05 '20 at 20:18
  • I edited the question but I am confused about it. It differs because we are actually changing the meaning of the first element. It's related to probability and actually the initial multiplication cannot work because it offsets the rest of the list. – DaraJ Dec 05 '20 at 20:51
  • But what they "should be" clearly depends on the meanings that come from the context - the "larger problem". If you don't know what any of the numbers should be we can't help. All that "related to probability" suggests is that the values sum to $1$. – Ethan Bolker Dec 05 '20 at 20:56
  • I'll try to make another example where I know the answer (n=2 still though). I actually didn't explain it the functions on it correctly, I'll completely re-write the question. – DaraJ Dec 05 '20 at 20:58
  • @EthanBolker Is that better? I am quite confused about this. – DaraJ Dec 05 '20 at 21:07
  • If the answer from @shortmanikos does not work for you we need an example with $n>2$ and at least three non-$0$ entries. If you can't tell us what that "should be" then I am afraid we can't help you. – Ethan Bolker Dec 05 '20 at 21:55
  • The answer @shortmanikos I think is correct, but the step I suspected to be wrong seems to be in fact wrong, which might leave me with a different question. Should I make a new question? – DaraJ Dec 05 '20 at 21:57
  • I have asked a new question: https://math.stackexchange.com/questions/3936321/changing-the-probabilities-of-state-transitions – DaraJ Dec 05 '20 at 22:05

1 Answers1

1

I am taking quite the long shot here, but how about

$ p_j' = (1-p_1') \cdot \dfrac{p_j}{\displaystyle\sum_{k=2}^n p_k} $,

where $ p_1, \dots , p_n $ are the original probabilities and $ p_j' $ the new ones. It keeps the sum at 1 and the "extra" probability is divided somewhat analogously.

shortmanikos
  • 1,425
  • Your method is correct I think and I'm using it now. But still my initial step ($\frac 1 2 -\left(\frac 1 2 \times \frac 2 7\right)$) is wrong – DaraJ Dec 05 '20 at 21:55