I What Is the Probability Atom A Will Emit a Photon Before Atom B?

Capitano
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Probability that one random gaussian event will happen before another one.
For concretness I'll use atoms and photons but this problem is actually just about probabilities.

There's an atom A whose probability to emit a photon between times t and t+dt is given by a gaussian distribution probability P_A centered around time T_A with variance V_A. There's a similar atom B described by a gaussian distribution P_B, but centered around T_B with variance V_B. Once they emit one photon, the process stops. What is the probability atom A will emit a photon before atom B? My attempt was something like this:

First, I ask a slightly different question. I start with the probability that A will emit a photon and B will not, between t and t+dt. That should be just

(P_A) x (1-P_B) x dt

since we require that A emits during that interval but not B. Now, my first idea now is to just integrate this expression from -infty to +infty, but I feel that's like demanding that in order for P_A to emit at some time, P_B needs to never emit during the whole time, which is not necessary. Another idea was to integrate up to a time t_f and then integrate over that time to infinity, but I'm not sure about that either
 
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Assuming that the decays are independent of each other then you can write the joint probability distribution as $$P(t_A,t_B)= P_A(t_A)\ P_B(t_B)$$ So then to calculate the probability that ##t_A<t_B## we integrate over the diagonal half plane where ##t_A<t_B## as follows $$ \int_{-\infty}^{t_B}P(t_A,t_B)\ dt_A$$ or equivalently $$ \int_{t_A}^{\infty}P(t_A,t_B)\ dt_B$$
 
Thanks for your answer! I am a bit confused about the free parameter that remains on the integral. For example, on your first expression, the result depends on t_B in the end. Does that mean "the probability that event A will happen first in the time interval from -infty to t_B"? I'd like to find out what's the probability that A happens first, without any free parameters in the end, just considering all the time interval. Is that possible?

Thanks again!
 
I am a little confused by some things in your problem statement, but here is my two cents based on how I interpreted your question:
##t_A-t_B## is a Gausian random variable with a mean ##T_{t_A}-T_{t_B}## (I think that it is the times that are random variables with a mean and variance) and a variance ##V_{t_A} + V_{t_B} + 2 cov(t_A,t_B)##. I assume that the random variables, ##t_A## and ##t_B## are independent, so ##cov(t_A,t_B)=0##. You are asking for the probability that ##t_A-t_B \lt 0##. You should be able to use the normal distribution tables to look up the probabilities you need.
 
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Capitano said:
I'd like to find out what's the probability that A happens first, without any free parameters in the end, just considering all the time interval. Is that possible?
Oops, you are right. I forgot the second integral:

integrate over the diagonal half plane where ##t_A<t_B## as follows $$ \int_{-\infty}^{\infty} \left( \int_{-\infty}^{t_B}P(t_A,t_B)\ dt_A \right) dt_B$$ or equivalently $$ \int_{-\infty}^{\infty} \left(\int_{t_A}^{\infty}P(t_A,t_B)\ dt_B \right) dt_A$$

For two normally distributed random variables the result should be as @FactChecker describes
 
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