Let $\Omega = (0,1], \mathcal{F} = \mathcal{B}(\Omega),P = \lambda$, where $\lambda$ is the Lebesgue measure with random variables, $$X_{n} := \sqrt{n}1_{\{\omega < 1/n\}}$$ $$Y_{n} := \frac{1}{2\sqrt{n\omega}}$$
Find: a) $E(X_{n})$
b)$E(Y_{n})$
c)$P(X_{n}\geq10)$
d)$P(Y_{n}\geq10)$
e)$P(Y_{n}\geq X_{n})$
My approach so far,
a) Since, we have the indicator function $$E(X_{n}) = E(\sqrt{n}1_{\{\omega < 1/n\}}) = \sqrt{n}E(1_{\{\omega < 1/n\}}) = \sqrt{n}P(\omega < 1/n) = \sqrt{n}/n = 1/\sqrt{n}.$$ Is the correct?
b)$$E(Y_{n}) = \int_{0}^{1}Y_{n}d\lambda(\omega) = \int_{0}^{1}\frac{1}{2\sqrt{n\omega}}d\lambda(\omega) = \int_{0}^{1}\frac{1}{2\sqrt{n\omega}}\lambda(d\omega) = \int_{0}^{1}\frac{1}{2\sqrt{n\omega}} d\omega = 1/\sqrt{n}.$$
Is this correct? I have used $\lambda(d\omega) = d\omega$ because $\lambda$ is the Lebesgue measure.
c) $P(X_{n}\geq10)$ = $P(\sqrt{n}1_{\{\omega < 1/n\}}\geq10) = P(1_{\{\omega < 1/n\}}\geq10/\sqrt{n}) = ?.$ How do I go about it from here? Do I take the intersection of both the events?
d) $P(Y_{n}\geq10)$ = $P\Big(\frac{1}{2\sqrt{n\omega}}\geq10\Big) = P\Big(\frac{1}{\sqrt{\omega}} \geq 20\sqrt{n} \Big) = P\Big({\sqrt{\omega}} \geq \frac{1}{20\sqrt{n}}\Big) = P\Big(\omega \leq \frac{1}{400n} \Big) = \frac{1}{400n}.$ Is this correct?
e) $P(Y_{n}\geq X_{n}) = P(Y_{n} - X_{n}\geq 0) = ?$ I tried replacing $X_{n}$ and $Y_{n}$ with what's given in the question but could not go further.
Any help will be appreciated.