I'm working with some probability distributions, working towards figuring the outcome of a statistical process.
I have worked out my survey results, and I have a confidence interval (Confidence Level 0.95) established, along with a standard deviation.
Now, I'm trying to work out the following. I've phrased it as clearly as I can, so it probably reads like a university exam question.
Let A(x) and B(x) be normal distributions with the same standard deviation, but different mean values.
Let $m$ represent the lower limit of $A$, and $n$ represent the upper limit of $A$, which represent the confidence interval (ie the actual result will be between them 95% of the time). The events are independent of each other.
In the same way, let $j$ and $k$ represent the lower and upper limits of $B$ respectively, where $j < m < k < n$, meaning the two distributions overlap.
ie:
$A=N(\sigma,\mu_{1})$,
$B=N(\sigma,\mu_{2})$
where $\mu_{2} < \mu_{1}$
What is the probability that the eventual result of the process where B prevails over A?
ie: We end up with two points, $x_{1}\epsilon A$ and $x_{2}\epsilon B$, is such that $x_{1} \leq x_{2}$ ?
My gut feeling is it is this:
$\Large P \normalsize _{(x_{1} \leq x_{2})} = \Large \int_j^m[\normalsize B(x_{2}) * \int_m^k \! A(x_{1}) \mathrm dx \Large] \mathrm dx$
Am I on the right track here? What's the correct formula?
Thanks in advance!
Since: $\mathsf P(B\leq A) = \mathsf P(A-B \geq 0)$
Then you wish to find the distribution of $(A-B)$ which is the sum of two Normally distributed random variables.
If $A\sim\mathcal{N}(\mu_{A}, \sigma_{A}^2)$ and $B\sim\mathcal{N}(\mu_{B}, \sigma_{B}^2)$ are independent, then: $$(A-B)\sim\mathcal N(\mu_{A}-\mu_{B}, \sigma_{A}^2+\sigma_{B}^2)$$
If the random variables are correlated, then we need to know the correlation coefficient $\rho$, and then you will have: $$(A-B)\sim\mathcal N(\mu_{A}-\mu_{B}, \sigma_{A}^2+\sigma_{B}^2-\rho_{{A},{B}}\;\sigma_{A}\;\sigma_{B})$$
Once you know the mean and variance of the difference, you can find: $\mathsf P(\frac{(A-B)-(\mu_{A-B})}{\sigma_{A-B}}>0)$