If I have two Gaussian distribution with the same $\sigma$ but different $\mu_x$ and $\mu_y$. How to calculate the $P(x>y\ |\ \mu_x\leq\mu_y)$? I think it's a type 1 error when I set the event "$\mu_x>\mu_y$" as null hypothesis?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Another way to interperate is:what is the minimum difference between $\mu_x$ and $\mu_y$ so that we can conclude they are significantly different? So should I calculate the distribution $z = x-y$ to see if the confidence intervals of z exlude 0? $\endgroup$– dexter2406Apr 6, 2020 at 19:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Since $\mu_x,\mu_y$ are constants (just like $\sigma$. If they're not, provide their distributions), the event in the given side has no additional information. It's like $1\leq 2$. So, $P(X>Y|\mu_x\leq \mu_y)=P(X>Y)$.
You'd probably wonder "if mean of x is less than y, the probability of X being larger than Y should be smaller, so it actually has information". This is already reflected in the final formula.
$T=X-Y$ will be another Gaussian RV with mean $\mu_x-\mu_y$ and variance $2\sigma^2$ if assumed independent. If $X$ and $Y$ are dependent, you need to provide the joint distribution. Then,
$$P(X-Y>0)=P(T>0)=P\left(Z>\frac{0-(\mu_x-\mu_y)}{\sqrt2\sigma}\right)=\Phi\left(\frac{\mu_x-\mu_y}{\sqrt 2\sigma}\right)$$
Where $\Phi(x)$ is the CDF of standard Gaussian RV. See that if $\mu_x\leq\mu_y$, the expression inside $\Phi$ will be negative and the probability will be smaller than $0.5$, while if $\mu_x\geq \mu_y$, it'll be positive and the probability will be larger than $0.5$.