Say we got a random variable $X$ with the Bernoulli distribution and parameter $\frac{1}{2}$, and a random variable $Y$ with the normal distribution with expectation $2(X-\frac{1}{2})$ and variance 1.
If I want to calculate $\mathbb P(X=1,Y\geq1)$, I know that I get 1 as expectation for $Y$; $$ \int_1^\infty \frac{1}{\sqrt{2\pi}}\operatorname{exp}(-\frac{1}{2}(y-1))\,\mathrm dy. $$ But what if I want to calculate $\mathbb P(Y\geq1)$?
Can I just fill in $X$ as a constant in the integral, because we don't integrate over $x$?; $$ \int_1^\infty \frac{1}{\sqrt{2\pi}}\operatorname{exp}\big(-\frac{1}{2}(y-2(X-\frac{1}{2})\big)\,\mathrm dy.$$ I need the form $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^1 \operatorname{exp}(-s^2/2)\,\mathrm ds$ for an exercise, so am I correct to substitute $y-2(X-\frac{1}{2})$ by $s$?
It just look weird to me to have $X$ in the integral...
Or should I write down $x$ instead of $X$, to indicate that we've chosen some arbitrary value that $X$ takes? Or should we also integrate over $x$? But that's not possible because $X$ is discrete...