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• This occurs when we select for a given $X$ a confidence interval for $\theta$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the possible cases of $\theta$ in the confidence interval (based on $L(\theta)$ and $U(\theta)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.
• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $\theta$ inside the interval and in a fraction $\alpha$ of the cases we will not include $\theta$ inside the interval.

(this is depicted in the image by the colored lines for the case $\theta=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

• This occurs when we select for a given $X$ a confidence interval for $\theta$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the possible cases of $\theta$ in the confidence interval (based on $L(\theta)$ and $U(\theta)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.

• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $\theta$ inside the interval and in a fraction $\alpha$ of the cases we will not include $\theta$ inside the interval.

  (this is depicted in the image by the colored lines for the case $\theta=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

Possibly a Clopper-Pearson interval may help to obtain an intuition about these confidence intervals. (the below is a variation of an answer to How to estimate a probability of an event to occur based on its count? more specifically it is a variation of a graph from Clopper-Pearson )

The main trick here is that we can switch L and U from being functions of X to being functions of $\mathbb{\theta}$

Imagine the case of 100 Bernoulli trials where the probability of success is $\theta$ and we observe the total number of successes $X$.

When we observe an $X$ as if it came from the unknown population of Bernoulli trials with true (unknown) probability $\theta$ then we will choose $U(X)$ and $L(X)$ such that no matter what the the real $\theta$ the probability to make a mistake is $\alpha$ in estimating $U(X)$ and L(X).

• This occurs when we select for a given $X$ a confidence interval for $\theta$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the possible cases of $\theta$ in the confidence interval (based on $L(\theta)$ and $U(\theta)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.
• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $\theta$ inside the interval and in a fraction $\alpha$ of the cases we will not include $\theta$ inside the interval.

(this is depicted in the image by the colored lines for the case $\theta=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

More formally:

if we choose a confidence interval such that

$$I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$$

then we have a $1-\alpha$ confidence interval.

The above means that we choose, for a given observation $x$, those $\theta$ into the interval for which the observation $x$ would occur within a $1-\alpha$ interval $P( L<x<U \vert \theta) = 1-\alpha$, where the $L$ and $U$ are now functions of $\theta$.

Then given any real $\theta$ we will observe:

• a fraction $1 - \alpha$ of the time an $X$ such that $$F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta)$$
• and a fraction $\alpha$ of the time an $X$ such that $$X < F_X(\alpha/2,\theta) \text{ or } X > F_X(1-\alpha/2,\theta)$$

If you like you could write out the $f_{L,U|\theta}$ which relates to $f_{X|\theta}$ and L(X) and U(X), both functions of X, are indeed correlated. But the figure above already shows enough, e.g. if $\theta=0.20$ then in 0+1+3+12+45+140 cases the $\theta<L(X)$ and in 196+48+8+1+0 cases the $\theta>U(x)$ while in 358+755+1297+1795+1974+1697+1119+551 cases $L(X) \leq \theta \leq U(x)$

If the shape of $x_U=U(\theta)$ and $x_L=L(\theta)$ is convex (like in the figure above), then we can use the inverses of those functions.

Possibly a Clopper-Pearson interval may help to obtain an intuition about these confidence intervals. (the below is a variation of an answer to How to estimate a probability of an event to occur based on its count? more specifically it is a variation of a graph from Clopper-Pearson )

The main trick here is that we can switch L and U from being functions of X to being functions of $\mathbb{\theta}$

Imagine the case of 100 Bernoulli trials where the probability of success is $\theta$ and we observe the total number of successes $X$.

When we observe an $X$ as if it came from the unknown population of Bernoulli trials with true (unknown) probability $\theta$ then we will choose $U(X)$ and $L(X)$ such that no matter what the the real $\theta$ the probability to make a mistake is $\alpha$ in estimating $U(X)$ and L(X).

• This occurs when we select for a given $X$ a confidence interval for $\theta$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the possible cases of $\theta$ in the confidence interval (based on $L(\theta)$ and $U(\theta)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.
• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $\theta$ inside the interval and in a fraction $\alpha$ of the cases we will not include $\theta$ inside the interval.

(this is depicted in the image by the colored lines for the case $\theta=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

$$-----------------------------------$$

More formally:

if we choose a confidence interval such that

$$I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$$

then we have a $1-\alpha$ confidence interval.

The above means that we choose, for a given observation $x$, those $\theta$ into the interval for which the observation $x$ would occur within a $1-\alpha$ interval $P( L<x<U \vert \theta) = 1-\alpha$, where the $L$ and $U$ are now functions of $\theta$.

Then given any real $\theta$ we will observe:

• a fraction $1 - \alpha$ of the time an $X$ such that $$F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta)$$
• and a fraction $\alpha$ of the time an $X$ such that $$X < F_X(\alpha/2,\theta) \text{ or } X > F_X(1-\alpha/2,\theta)$$

If you like you could write out the $f_{L,U|\theta}$ which relates to $f_{X|\theta}$ and L(X) and U(X) (both L(X) and U(X), functions of X, are indeed correlated). 

But the figure above already shows enough, e.g. if $\theta=0.20$ then in 0+1+3+12+45+140 cases the $\theta<L(X)$ and in 196+48+8+1+0 cases the $\theta>U(X)$ while in 358+755+1297+1795+1974+1697+1119+551 cases $L(X) \leq \theta \leq U(X)$

$$-----------------------------------$$

If the shape of $x_U=U(\theta)$ and $x_L=L(\theta)$ is convex (like in the figure above), then we can use the inverses of those functions.

Possibly a Clopper-Pearson interval may help to obtain an intuition about these confidence intervals. (the below is a variation of an answer to How to estimate a probability of an event to occur based on its count? more specifically it is a variation of a graph from Clopper-Pearson )

The main trick here is that we can switch L and U from being functions of X to being functions of $\mathbb{\theta}$ (or $\theta=p$ in the particular case of Bernoulli trials, that is used in this answer)

Imagine the case of 100 Bernoulli trials where the probability of success is $p$ and we observe the total number of successes $X$.

When we observe an $X$ as if it came from the unknown population of Bernoulli trials with true probability $p$ then we will choose $U(X)$ and $L(X)$ such that no matter what the the real $p$ (or $\theta$) the probability to make a mistake is $\alpha$ in estimating $U(X)$ and L(X).

• This occurs when we select for a given $X$ a confidence interval for $p$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the cases of $p$ in the confidence interval (based on $L(p)$ and $U(p)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.
• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $p$ inside the interval and in a fraction $\alpha$ of the cases we will not include $p$ inside the interval.

(this is depicted in the image by the colored lines for the case $p=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

More formally:

if we choose a confidence interval such that

$$I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$$

then we have a $1-\alpha$ confidence interval.

The above means that we choose, for a given observation $x$, those $\theta$ into the interval for which the observation occurs within a $1-\alpha$ interval $P( L<x<U \vert \theta) = 1-\alpha$, in which the $L$ and $U$ are now functions of $\theta$ (or $p$).

Then given any real $\theta$ or $p$ we will observe:

• a fraction $1 - \alpha$ of the time an $X$ such that $$F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta)$$
• and a fraction $\alpha$ of the time an $X$ such that $$X < F_X(\alpha/2,\theta) \text{ or } X > F_X(1-\alpha/2,\theta)$$

If you like you could write out the $f_{L,U|\theta}$ which relates to $f_{X|\theta}$ and L(X) and U(X), both functions of X, are indeed correlated. But the figure above already shows enough, e.g. if p=20 then in 0+1+3+12+45+140 cases the $\theta<L(X)$ and in 196+48+8+1+0 cases the $\theta>U(x)$ while in 358+755+1297+1795+1974+1697+1119+551 cases $L(X) \leq \theta \leq U(x)$

If the shape of $x_U=U(\theta)$ and $x_L=L(\theta)$ is convex (like in the figure above), then we can use the inverses of those functions.

Possibly a Clopper-Pearson interval may help to obtain an intuition about these confidence intervals. (the below is a variation of an answer to How to estimate a probability of an event to occur based on its count? more specifically it is a variation of a graph from Clopper-Pearson )

The main trick here is that we can switch L and U from being functions of X to being functions of $\mathbb{\theta}$

Imagine the case of 100 Bernoulli trials where the probability of success is $\theta$ and we observe the total number of successes $X$.

When we observe an $X$ as if it came from the unknown population of Bernoulli trials with true (unknown) probability $\theta$ then we will choose $U(X)$ and $L(X)$ such that no matter what the the real $\theta$ the probability to make a mistake is $\alpha$ in estimating $U(X)$ and L(X).

• This occurs when we select for a given $X$ a confidence interval for $\theta$ (based on $L(x)$ and $U(x)$) such that $X$ occurs in a fraction $1-\alpha$ of all the possible cases of $\theta$ in the confidence interval (based on $L(\theta)$ and $U(\theta)$). There is some degree of freedom in shifting more or less weight between $U$ and $L$ and there are many different ways to do this.
• If we do this consistently every-time that we perform an experiment, then in a fraction $1-\alpha$ of the cases we will observe an $X$ that let's us include the true $\theta$ inside the interval and in a fraction $\alpha$ of the cases we will not include $\theta$ inside the interval.

(this is depicted in the image by the colored lines for the case $\theta=0.2$, the gray lines are cases when we select the right interval, the red when the interval is too high and the green when the interval is too low.)

More formally:

if we choose a confidence interval such that

$$I_{\alpha}(X) = \lbrace \theta: F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta) \rbrace$$

then we have a $1-\alpha$ confidence interval.

The above means that we choose, for a given observation $x$, those $\theta$ into the interval for which the observation $x$ would occur within a $1-\alpha$ interval $P( L<x<U \vert \theta) = 1-\alpha$, where the $L$ and $U$ are now functions of $\theta$.

Then given any real $\theta$ we will observe:

• a fraction $1 - \alpha$ of the time an $X$ such that $$F_X(\alpha/2,\theta) \leq X \leq F_X(1-\alpha/2,\theta)$$
• and a fraction $\alpha$ of the time an $X$ such that $$X < F_X(\alpha/2,\theta) \text{ or } X > F_X(1-\alpha/2,\theta)$$

If you like you could write out the $f_{L,U|\theta}$ which relates to $f_{X|\theta}$ and L(X) and U(X), both functions of X, are indeed correlated. But the figure above already shows enough, e.g. if $\theta=0.20$ then in 0+1+3+12+45+140 cases the $\theta<L(X)$ and in 196+48+8+1+0 cases the $\theta>U(x)$ while in 358+755+1297+1795+1974+1697+1119+551 cases $L(X) \leq \theta \leq U(x)$

If the shape of $x_U=U(\theta)$ and $x_L=L(\theta)$ is convex (like in the figure above), then we can use the inverses of those functions.