Probability Question

I have been racking my brains trying to come up with a solution for the following,

I will use the financial markets as an example for this question as it seems easiest to illustrate in this way.

Gold is pricing at 1650.00, I need an equation for the probability of Gold trading 1660 and 1640 within a certain time. Let use 1 hour for the example.

Therefore the question I have is:

Find the probability of Gold trading both 1660 and 1640 within a 1 hour period. These events are mutually exclusive as Gold cannot be trading two prices at the same time, the probability of both A and B occuring are variable due to the market constantly moving up and down. If the market moves above 1660 then A has been satisfied, however B is now less likely to be satisifed within the time constraint. As the time diminishes the probability of both events occuring also diminishes.

If anyone has any insight into the formula for the above question it would be a great help.

Thanks in advance.

my 2 cents worth.

if you change your question to Gold touching just one spot, then this is just an option pricing question, and the BS option pricing model is your answer (although you may have to figure out a way to change the $ into % of probability)

Quote: andysif

my 2 cents worth.

if you change your question to Gold touching just one spot, then this is just an option pricing question, and the BS option pricing model is your answer (although you may have to figure out a way to change the $ into % of probability)

Yes it does make it much easier if we only need to look at the one touch, but it is this double touch that I am interested in.

P(A,B) = P(A|B)*P(B).
Homework?

slightly trickier than the average homework question!!

You are neglecting the fact that it can touch in either order, and that the probability of either occuring is constantly variable with both time to expiry and distance from the other point.

Quote: Dan2186

slightly trickier than the average homework question!!

I could be missing something, but it does not look like it to me ...
Besides, my comment was not only based on the fact that the question seems trivial, but also on that it does not look practically useful for anything (this is just a lay person first impression, because I have no knowledge of gold trading whatsoever).

Quote:

You are neglecting the fact that it can touch in either order,

Come on.
Fix the order, calculate once, then swap A and B, and do it again, then P = P1 + (1-P1)*P2

Quote:

and that the probability of either occuring is constantly variable with both time to expiry and distance from the other point.

Yes, you need to integrate it over the whole interval. I am still not seeing the complexity (provided that you know how to calculate P(A), P(B), P(A|B) and P(B|A) at a fixed point in time, which I have no idea how to do, but you have indicated, that it is a simple task).

Fear & Greed

Let me put some more context to it.

I am trying to price an options. With your above information I am fairly close to what my calculations have attained. However I am still arriving at rogue figures.

However using my earlier example of Strike A:1660 and Strike B: 1640 this is where I run into problems:

I would expect
I would expect that when the market is trading 1660, therefore Stike A has been hit that P(B)=0.4370 however with your equation I am arriving at P(B)=0.449348.
Also when the market goes through this first strike to say 1700 i'd expect P(B) to be approx 0.0208 which would be Gold having a $60 dollar move once every 50 days or so wheras with your equation we arrive at P(B)= 0.437555, which doesn't sit well. If I then say that Gold continues to rise to 1800 then the probability of coming back to 1640 should be near on zero, however I still arrive at 0.436964 with your equations.

Any other thoughts Weaselman?

Would I be wrong in thinking that this problem can be simplified to:

P(A and B)= P(A) * P(B)

P(A or B)= P(A) + P(B) - P(A and B)

your commodities broker, hard at work...

Quote: Dan2186

Would I be wrong in thinking that this problem can be simplified to:

P(A and B)= P(A) * P(B)

P(A or B)= P(A) + P(B) - P(A and B)

The second line is correct.
The first one would be true if A and B were independent, but, as you pointed out they are not (P(A|B) < P(A)), so, unless you are willing to ignore the correlation for simplicity, the correct formula is P(A,B) = P(A|B)*P(B).

Quote: Dan2186

I would expect that when the market is trading 1660, therefore Stike A has been hit that P(B)=0.4370 however with your equation I am arriving at P(B)=0.449348.
Also when the market goes through this first strike to say 1700 i'd expect P(B) to be approx 0.0208 which would be Gold having a $60 dollar move once every 50 days or so wheras with your equation we arrive at P(B)= 0.437555, which doesn't sit well. If I then say that Gold continues to rise to 1800 then the probability of coming back to 1640 should be near on zero, however I still arrive at 0.436964 with your equations.

I am afraid I know too little about the domain (which is a long way to say that I am completely ignorant) to be able to understand what you are talking about here - like why you expected the value to be what you said, etc.

If you can formulate the assumptions you are making in math terms, and show how you used them to arrive at the answer, I could try to be of more help.

Quote: Dan2186

Yes it does make it much easier if we only need to look at the one touch, but it is this double touch that I am interested in.

OK. Using your example, spot is 1650, then 1640 and 1660 is each x sd away from the spot. Then say the probability of reaching 1640 is P(x), and after that, the probability of reaching 1660 is P(2x), because its twice as far away(note: P(2x) =\= 2*P(x)). Therefore the probability of reach 1640 then 1660 is P(x) * P(2x), and then you double that for 1660 then 1640, and it should be you answer.