Liam Keller

2022-09-17

Confidence interval multiplication

The question looks pretty simple but I can't get my hands on it:Say I have a probability which is the product of two other independent probabilities $p={p}_{1}{p}_{2}$.

I have estimated each probability ${p}_{1}$ and ${p}_{2}$ and found some 95% confidence interval for each. How do I obtain a 95% confidence interval for p?

Taking the product of the bounds of the interval won't work as I would be taking 95% of a 95% confidence interval resulting in an approximately 90% which is not what I want.So conversely I would be taking 97.5% confidence interval for each and by multiplying the bounds I will obtain a 95% confidence interval, is that right?

I feel like something is going wrong. In my situation I deal with probabilities but it could be anything so this question can be generalised to any type of confidence intervals.

If my reasoning is correct, could someone convince me that's the correct way of doing so?

The question looks pretty simple but I can't get my hands on it:Say I have a probability which is the product of two other independent probabilities $p={p}_{1}{p}_{2}$.

I have estimated each probability ${p}_{1}$ and ${p}_{2}$ and found some 95% confidence interval for each. How do I obtain a 95% confidence interval for p?

Taking the product of the bounds of the interval won't work as I would be taking 95% of a 95% confidence interval resulting in an approximately 90% which is not what I want.So conversely I would be taking 97.5% confidence interval for each and by multiplying the bounds I will obtain a 95% confidence interval, is that right?

I feel like something is going wrong. In my situation I deal with probabilities but it could be anything so this question can be generalised to any type of confidence intervals.

If my reasoning is correct, could someone convince me that's the correct way of doing so?

Zayden Dorsey

Beginner2022-09-18Added 18 answers

Step 1

I can give you a partial answer, which is you are on the right track, but multiplying the 97.5% bounds will not yield a 95% confidence interval. Your interval will be too large.

I think what you are trying to do is estimate two unknown proportions ${p}_{1}$ and ${p}_{2}$. You compute two estimates ${\hat{p}}_{1}$ and ${\hat{p}}_{2}$ from the data you observe. You compute confidence intervals for each one ${I}_{1}=[{a}_{1},{b}_{1}]$ and ${I}_{2}=[{a}_{2},{b}_{2}]$, respectively. So, $P({p}_{1}\in {I}_{1})=P({p}_{2}\in {I}_{2})=0.975$. Since, they are independent, you can consider the confidence set $S={I}_{1}\times {I}_{2}$, which, geometrically, is a square in the plane. Now, $P(({p}_{1},{p}_{2})\in S)=0.95$ since we have independence. This is equivalent to $P({p}_{1}\in {I}_{1},{p}_{2}\in {I}_{2})=P({p}_{1}\in {I}_{1})P({p}_{2}\in {I}_{2})$. That is, the probability of this ordered pair being inside the square S is 95%. This is a way to get that 95% confidence set.

Step 2

However, if you are interested in a confidence interval for $p={p}_{1}{p}_{2}$ multiplying the bounds together to get an interval $I=[{a}_{1}{a}_{2},{b}_{1}{b}_{2}]$ will cause a problem. Using the same notation, you are asking for the probability $P(p\in I)$. Putting some concrete numbers to this, let ${I}_{1}={I}_{2}=[0.5,0.6]$, so that $I=[0.25,0.36]$. All pairs $({p}_{1},{p}_{2})\in [0.5,0.6{]}^{2}$ yield a $p\in [0.25,0.36]$, so $P(p\in I)\ge 0.95$. The problem is that there are also pairs $({p}_{1},{p}_{2})\notin [0.5,0.6{]}^{2}$ that have $p\in [0.25,0.36]$ such as $({p}_{1},{p}_{2})=(0.3,1.0)$ where $p=0.3$. Hence, $P(p\in I)>0.95$. Your confidence interval is too big.

In general, it will require more effort to produce a confidence interval for the product ${p}_{1}{p}_{2}$. Even if the your data is coming from nice distributions (e.g. normal, binomial) the product distribution will probably be much nastier. Off the top of my head, I do not have a good answer to how to produce such an interval.

I can give you a partial answer, which is you are on the right track, but multiplying the 97.5% bounds will not yield a 95% confidence interval. Your interval will be too large.

I think what you are trying to do is estimate two unknown proportions ${p}_{1}$ and ${p}_{2}$. You compute two estimates ${\hat{p}}_{1}$ and ${\hat{p}}_{2}$ from the data you observe. You compute confidence intervals for each one ${I}_{1}=[{a}_{1},{b}_{1}]$ and ${I}_{2}=[{a}_{2},{b}_{2}]$, respectively. So, $P({p}_{1}\in {I}_{1})=P({p}_{2}\in {I}_{2})=0.975$. Since, they are independent, you can consider the confidence set $S={I}_{1}\times {I}_{2}$, which, geometrically, is a square in the plane. Now, $P(({p}_{1},{p}_{2})\in S)=0.95$ since we have independence. This is equivalent to $P({p}_{1}\in {I}_{1},{p}_{2}\in {I}_{2})=P({p}_{1}\in {I}_{1})P({p}_{2}\in {I}_{2})$. That is, the probability of this ordered pair being inside the square S is 95%. This is a way to get that 95% confidence set.

Step 2

However, if you are interested in a confidence interval for $p={p}_{1}{p}_{2}$ multiplying the bounds together to get an interval $I=[{a}_{1}{a}_{2},{b}_{1}{b}_{2}]$ will cause a problem. Using the same notation, you are asking for the probability $P(p\in I)$. Putting some concrete numbers to this, let ${I}_{1}={I}_{2}=[0.5,0.6]$, so that $I=[0.25,0.36]$. All pairs $({p}_{1},{p}_{2})\in [0.5,0.6{]}^{2}$ yield a $p\in [0.25,0.36]$, so $P(p\in I)\ge 0.95$. The problem is that there are also pairs $({p}_{1},{p}_{2})\notin [0.5,0.6{]}^{2}$ that have $p\in [0.25,0.36]$ such as $({p}_{1},{p}_{2})=(0.3,1.0)$ where $p=0.3$. Hence, $P(p\in I)>0.95$. Your confidence interval is too big.

In general, it will require more effort to produce a confidence interval for the product ${p}_{1}{p}_{2}$. Even if the your data is coming from nice distributions (e.g. normal, binomial) the product distribution will probably be much nastier. Off the top of my head, I do not have a good answer to how to produce such an interval.

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