# Confidence intervals on maximum likelihoods of observed data

I observed 400 episodes of nursing care in a hospital. I tracked the movement of the nurses between 5 rooms $A-E$. The maximum likelihood of them moving from room $i\rightarrow j$ is given by:

$$P_{ij}=\displaystyle \dfrac{\text{# of times from room i\rightarrow j}}{\displaystyle \text{Total # of transitions to any room}}$$

• Is there a way of defining a confidence interval on this maximum likelihood estimate $P_{ij}$?
• And for all maximum likelihood estimates of all possible room combinations?

Reference:

I have come across a reference: http://arxiv.org/pdf/0905.4131v1.pdf This suggests that for n observations $X_i$, the empirical maximum likelihood estimate $\hat P_{ij}$ minus the actual transition probability $P_{ij}$ would tend to a multivariate normal distribution with mean 0 and matrix of variance-covariances $\Sigma$.

$$\sqrt{n}|\hat{P_{ij}}-P_{ij}|\sim N(0,\Sigma)\quad \text{as}\quad n\rightarrow \infty$$

How to I calculate $\Sigma$ from my observed data? And how does this relate to confidence intervals?

• It would be a confidencce interval for the unobservable proportion, based on the observable MLE. Jun 28, 2013 at 20:54
• . . . also, one should call this a maximum likelihood estimate of that proportion. Jun 28, 2013 at 20:54
• Could you elaborate on that for me at all?
– HCAI
Jun 28, 2013 at 20:58

One could probably think about the nature of the dependence among the various random variables that would have been observed, but for now I'll do something simpler:

You have $n$ independent Bernoulli trials; in this case $n=400$. You have $x$ successes; in this case, $x$ is the numerator in the fraction. So the number of successes is binomially distributed with an unobservable parameter $p$. Theory tells us the expected value of the number of successes is $400p$ and the variance of the number of successes is $400p(1-p)$. That means the expected proportion of successes is $p$ and the variance of the proportion is $p(1-p)/400$. So we can cite the central limit theorem and we have an approximately normal distribution; thus about a $0.95$ chance of being between $-1.96$ and $+1.96$. We use $x/400$ as an estimate of $p$. Our $95\%$ confidence interval therefore has endpoints $$\frac{x}{400} \pm 1.96\sqrt{\frac{(x/400)(1-(x/400))}{400}}$$

In many textbooks, you'll see a section called something like "Confidence interval for a proportion" that covers this.

• Yes I think I see, so in this case $i\rightarrow j$ is considered a success and $i\rightarrow$ anywhereelse, a fail. AT the end of the day, these MLE form a stochastic matrix $P$, such that $P_{ij}$ is the probability of moving from $i\rightarrow j$. Then each $P_{ij}$= your CI formula? But by definition $\sum_{i=1..5}P_i=1$, how can one account for this with a CI?
– HCAI
Jun 28, 2013 at 21:42
• Not only is $i\to\text{anywhere else}$ a failure, but $\text{anywhere else}\to j$ is a failure and $\text{anywhere else}\to\text{anywhere else}$ is a failure. Jun 29, 2013 at 2:40
• I was deriving a confidence interval for only one $P_{ij}$. If one considered all of them, then one would have to account for the dependence among them. Jun 29, 2013 at 2:41
• So the above formula could be used for all room combinations (e.g. A-B,B-E,E-D, etc, in total 25) but would not reflect the dependence of these? What would be the method for accounting for dependence between them to calculate a CI?
– HCAI
Jun 29, 2013 at 19:53
• I understand that I would need to calculate the variance and covariance from the data as the MLE follow a multivariate normal distribution. But how? I only have one set of observations?
– HCAI
Jul 1, 2013 at 12:08