Properties and Applications of Probability Distributions I am an MBA Student taking courses in Statistics.
Our prof gave us a data manipulation assignment in which we have a column that contains a single weather measurement over a period of 100 days. For example, the weather can be either Sunny, Rainy, Cloudy, Snowy or Windy.
Based on this data, we have to calculate the probability of "two consecutive days of any weather combination". For example, what is the probability of a Sunny day followed by another Sunny day - or what is the probability of a Cloudy day followed by a Windy day?
As such, I think there are 25 possible combinations - I made a 5 x 5 contingency table in which each entry consists of a possible weather combination. Using the data, I calculated the probability of each possible weather combination and populated the entries within the table. As a logical check, I ensured that the sum of probabilities in any given row always add up to 1.
Although this was all the assignment was asking us for, I thought of the following idea: What if these actually aren't the "true" probabilities? By this I mean, what if this weather over these 100 days was unusual - is it possible to place a "range" on these probabilities?

*

*For example, the probability of two consecutive Sunny days is 0.13, but in reality this probability could be anywhere between 0.11 and 0.15.

*Or, given that today is Snowy, there is a 0.3 ± 0.05 probability that it will be Sunny tomorrow

This way, given the current weather, I could compare the probabilities and ranges for all possible weather combinations and find out the most likely weather for tomorrow.
I spent the day thinking about this question, and it sounds that maybe a "confidence interval" might be the answer I am looking for. Now, the question becomes how to calculate the confidence interval for these 25 probabilities.
I kept thinking about this question and thought that this question is like rolling a 5 sided dice. We learned that flipping a two sided coin is a Binomial Probability Distribution whereas anything with more than two sides is a Multinomial Probability Distribution (https://en.wikipedia.org/wiki/Multinomial_distribution).
I know that the general formula of a 95% confidence interval (based on the standard deviation of the Binomial Distribution) for a proportion can be given by :  Estimated Proportion ± 1.96 sqrt[ (Estimated Proportion * (1 - Estimated Proportion)) / sqrt(n) ].
In the case of the Multinomial Probability Distribution - could I say that the 95% confidence interval: Estimated Proportion ± sqrt[n * Estimated Proportion * (1-Estimated Proportion)]?
In the above formula, "n" would be the total number of points (i.e. 100) and I would repeat this formula for each of the 25 probabilities. This formula is based on the Standard Deviation of the Multinomial Distribution.
Is my understanding of this correct?
 A: The idea to construct confidence intervals can be a good idea. However, pay attention to the following facts.

*

*First let us recall some definitions: a $1-\alpha$ confidence interval for a parameter $\theta$ is an interval $C_n=(a, b)$ where $a=a\left(X_1, \ldots, X_n\right)$ and $b=b\left(X_1, \ldots, X_n\right)$ are functions of the data such that
$$
\mathbb{P}_\theta\left(\theta \in C_n\right) \geq 1-\alpha, \text { for all } \theta \in \Theta \text {. }
$$
In words, $(a, b)$ traps $\theta$ with probability $1-\alpha$. We call $1-\alpha$ the coverage of the confidence interval.
Pay attention to the fact that $C_n$ is random and $\theta$ is fixed. Why is this important? Because many people think that confidence intervals are probability statements about the truth $\theta$, but they are not! A confidence interval is not a probability statement about $\theta$ since $\theta$ is a fixed quantity, not a random variable. Some texts interpret confidence intervals as follows: if I repeat the experiment over and over, the interval will contain the parameter 95 percent of the time. This is correct but useless since we rarely repeat the same experiment over and over. A better interpretation is given by Larry Wasserman as follows: on day 1 , you collect data and construct a 95 percent confidence interval for a parameter $\theta_1$. On day 2 , you collect new data and construct a 95 percent confidence interval for an unrelated parameter $\theta_2$. On day 3, you collect new data and construct a 95 percent confidence interval for an unrelated parameter $\theta_3$. You continue this way constructing confidence intervals for a sequence of unrelated parameters $\theta_1, \theta_2, \ldots$ Then 95 percent of your intervals will trap the true parameter value. There is no need to introduce the idea of repeating the same experiment over and over.

Having said this, what I mean is that if you construct a confidence interval actually you will know little about the truth $\theta$, you would know that if you were to repeat the recollection of data during those 100 days for 100 consecutive years, and each year you were to construct a confidence interval for your estimated probability of interest, then the confidence intervals would contain the true $\theta$ about 95 years out of the 100 years you recollected the data.

*

*The confidence interval that you propose is not an exact confidence interval to see that, compare the following. Example: in the coin flipping setting, let $C_n=\left(\widehat{p}_n-\epsilon_n, \widehat{p}_n+\epsilon_n\right)$ where $\epsilon_n^2=\log (2 / \alpha) /(2 n)$. From Hoeffding's inequality it follows that
$$
\mathbb{P}\left(p \in C_n\right) \geq 1-\alpha
$$
for every $p$. Hence, $C_n$ is a $1-\alpha$ confidence interval. Let now $X_1, \ldots, X_n \sim \operatorname{Bernoulli}(p)$ and let $\widehat{p}_n=n^{-1} \sum_{i=1}^n X_i$. Then
$$\mathbb{V}\left(\widehat{p}_n\right)=n^{-2} \sum_{i=1}^n \mathbb{V}\left(X_i\right)=n^{-2} \sum_{i=1}^n p(1-p)=n^{-2} n p(1-p)=p(1-p) / n$$
Hence, se $=\sqrt{p(1-p) / n}$ and $\widehat{\mathrm{se}}=\sqrt{\widehat{p}_n\left(1-\widehat{p}_n\right) / n}$. By the Central Limit Theorem, $\widehat{p}_n \approx N\left(p, \widehat{\text { se }}^2\right)$. Therefore, an approximate $1-\alpha$ confidence interval is
$$
\widehat{p}_n \pm z_{\alpha / 2} \widehat{\mathrm{se}}=\widehat{p}_n \pm z_{\alpha / 2} \sqrt{\frac{\widehat{p}_n\left(1-\widehat{p}_n\right)}{n}} .
$$
Compare this with the confidence interval above. The Normal-based interval is shorter but it only has approximately (large sample) correct coverage.

In your proposed confidence interval, you are doing as in the second part of the example where you get an approximate confidence interval, not an exact one. So you would have to account also for uncertainty due to the small sample size.

*

*Your idea of modeling the problem as a 5-sided dice is nice, but is not exactly correct. Because in your model you would be throwing the dice twice consecutively. Therefore some adjustment is needed if you want to use the confidence interval for the Multinomial (i.e. you cannot use the parameters you proposed).

Having said so, I hope this lengthy answer will help you to make a correct decision regarding which statistical tool you need and if indeed it is really the confidence interval the one you want.
Main source of the answer: Larry Wasserman - Lecture Notes
