# Find probability that confidence intervals are disjoint

Given $X_1, \ldots, X_{100}$ and $Y_1,\ldots,Y_{50}$ which are independent random samples from identical distribution $N(\mu,1)$. Each of two statisticians is trying to build confidence interval for $\mu$ on a confidence level of $0.8$ ($1-\alpha=0.8$). You have to find probability that those intervals are disjoint. I know i have to find $P(|\overline{X}-\overline{Y}|> \frac{1\cdot1.28}{\sqrt{50}} + \frac{1\cdot1.28}{\sqrt{100}})$ which is $1-P(-0.309 \le \overline{X}-\overline{Y}\le 0.309)$

Question is what should i do next how to calculate this probability?

EDIT: Thus the solution is $1-(\phi(1.28)-(1-\phi(1.28)))\approx0.2$?

• $\bar X$ is $N(\mu, 1/\sqrt{100})$ and $\bar Y$ is $N(\mu, 1/\sqrt{50})$. $\bar X - \bar Y$ is a linear combination of independent normals which is also normal. $\bar X - \bar Y$ has mean $\mu - \mu = 0$. Find the standard deviation of $\bar X - \bar Y$ and then you can complete the computation. Commented Aug 6, 2017 at 18:32
• and than simple standardization? Commented Aug 6, 2017 at 18:35
• Care to elaborate on how you get 2 from $1/\sqrt{100}$ and $1/\sqrt{50}$? Commented Aug 6, 2017 at 18:37
• my mistake, i was thinking about something different Commented Aug 6, 2017 at 18:38
• added something to the main question Commented Aug 6, 2017 at 18:45

One interval has endpoints $\bar X \pm 1.28 \dfrac 1 {\sqrt{100}}$ and the other $\bar Y \pm 1.28 \dfrac1 {\sqrt{50}}.$
The intervals are disjoint if $\bar X + 1.28 \dfrac 1 {\sqrt{100}} < \bar Y - 1.28 \dfrac 1 {\sqrt{50}}$ or $\bar Y + 1.28 \dfrac 1 {\sqrt{50}} < \bar X - 1.28 \dfrac 1 {\sqrt{100}}.$
That happens if $\bar Y - \bar X < 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right)$ or $\bar X - \bar Y > 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right).$ In other words $$|\bar X - \bar Y| > 1.28 \left( \dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}} \right).$$ This is equivalent to $$\frac{|\bar X - \bar Y|}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} > 1.28.$$
You have $\bar X - \bar Y \sim N\left(0, \dfrac 1 {100} + \dfrac 1 {50} \right) = N\left( 0, \dfrac 3 {100} \right),$ so that $$\frac{\bar X - \bar Y}{ \sqrt{3/100} } \sim N(0,1).$$ \begin{align} & \operatorname{standard deviation}\left( \frac{\bar X - \bar Y}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} \right) = \frac{\operatorname{standard deviation}\left( \bar X - \bar Y \right)}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} \\[10pt] = {} & \dfrac{\sqrt{\dfrac 3 {100}}}{\dfrac 1 {\sqrt{100}} + \dfrac 1 {\sqrt{50}}} = \frac{\sqrt 3}{1 + \sqrt 2} = \sqrt 6 - \sqrt 3. \end{align} So \begin{align} & \Pr\left( \frac{|\bar X - \bar Y|}{\sqrt{\dfrac 1{100}} + \sqrt{\dfrac 1 {50}}} > 1.28 \right) = \Pr\left( \left. \frac{|\bar X - \bar Y|}{\sqrt{\dfrac 1{100}} + \sqrt{\dfrac 1 {50}}} \right/ (\sqrt6-\sqrt3) > \frac{1.28}{\sqrt6-\sqrt3} \right) \\[10pt] = {} & \Pr\left( |Z| > \frac{1.28}{\sqrt6-\sqrt 3} \right) \text{ where } Z\sim N(0,1). \end{align}