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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...

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    There's not a closed form for this answer; the integral can't be done this way. – 2017-01-31
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    $P(Y≥1)=P(Y≥1\, | \, X=0)P(X=0)+ P(Y≥1\, | \, X=1)P(X=1)$ – 2017-01-31
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    @Paul I'm very happy with lulu's solution, but why wasn't my approach correct? We know that $X$ will take some value, say $x$, and if we integrate over $y$, it doesn't matter what that value is, right? Edit; Ohh, I think I understand. $X$ is of course a function of $\omega$, so if we want to determine $\mathbb P$, we have to sort of fill in the $\omega$. – 2017-01-31
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    That's fine if the form lulu gives is adequate, but it can't be found exactly. – 2017-01-31

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