# Can I use another parameter in an unbiased estimator?

Given a multinomial distribution, the task is to make an unbiased estimator for one parameter. However, the parameters all add up to 1, so how can I make an unbiased estimator without using other parameters. Am I allowed to use other parameters?

• Can you explain your problem in a more detailed manner? Your question seems very general for us to help you Commented May 1, 2021 at 20:18
• So there is an event (X(i)) that can either succeed, fail or neither. The event occurs n times. X(i) = 0 if failed, 1 if neither and 2 if success. The parameters for the probability of each event occurring is labeled L1, L2, L3. So L1+L2+L3=1. I need to come up with an unbiased estimator for L1. Commented May 1, 2021 at 20:25
• Thanks. Think about a binomial event for simplicity. Probabilities are $p$ and $q$, respectively. Then $p+q=1$. In principle if you know $q$ then you know $p$, and using that the expectation is linear then given an unbiased estimator for $q$ you would have an unbiased estimator for $p$ as well. In that sense, $p$ and $q$ are not independent. The same happens with your multinomial distribution Commented May 1, 2021 at 20:27
• I see what you mean but with binomial I can simply adjust the event results so that it is bernoulli and then the sample mean will give me an estimator, but I don't know how to do that here. Commented May 1, 2021 at 20:29
• Just relabel the events as $X=0$ and $X\neq 0$ Commented May 1, 2021 at 20:30

Think about a binomial for simplicity. Probabilities are $$p$$ and $$q$$, respectively. Then $$p+q=1$$. In principle if you know $$q$$ then you know $$p$$, and using that the expectation is linear then given an unbiased estimator for $$q$$ you would have an unbiased estimator for $$p$$ as well. In that sense, $$p$$ and $$q$$ are not independent. The same happens with your multinomial distribution. In that case an unbiased estimator for $$L_1$$ would be the sample proportion of observations where $$X=1$$. That is
$$\hat{L}_1=\sum 1_{X(i)=0}/n$$
Notice that $$1_{X(i)=0}$$ has a Bernoulli distribution with probability $$L_1$$ and the rest follows from there.
• I am not sure, you probably can do it "without conditionals". However, notice that since you are working with a Bernoulli RV the variance is just $\hat{L}_1 *(1-\hat{L}_1)$. Commented May 1, 2021 at 20:58