# How would I calculate the bias of the estimator Û=(X₂+X₃+…+Xn)/(N+10)

The observations $X_1,\dots, X_n$ form an IID sample from a probability distribution with unknown mean $U$. Let

$$\hat{U}=\frac{X_2+X_3+\dots+X_n}{n+10}$$

be an estimator used to estimate $U$. (I.e., compared to the more usual estimator $\bar{X}$, $\hat{U}$ discards the first observation and uses the divisor $n+10$ rather than $n$). The question I want to answer is what is the bias of $\hat{U}$?

I am aware that the bias ($b$) of an estimator ($\hat{U}$) for $U$ is

$$b=E(\hat{U})-U$$,

that is, the bias is equal to the expectation of the estimator minus the actual value of the estimand. In this case

$$b=E\bigg(\frac{X_2+X_3+\dots+X_n}{n+10}\bigg)-U.$$

I know that I cannot get a specific value but the question I have been given has asked me to simplify this question down and I am unsure how.

Also in an example given it states that the bias of $\hat{U}$ is $-U/(N+1)$, but I am unclear how to get this answer.

• What are you trying to estimate? In your formula for $b$ and in the title you begin indexing with $2$: is that on purpose or is it a typographical error? – whuber Jan 4 '18 at 14:51
• Can you clarify what U represents in this problem? Also, do we know the expected value of the $X_i$s? Do they come from the same distribution? – Anshu says Reinstate Monica Jan 4 '18 at 14:54
• Thank you. When you incorporate that information in your post, it will come to the attention of the community an enable people to vote to reopen it. – whuber Jan 4 '18 at 17:04
• Thank you for the edit. I have tried to improve the post a little more and voted to reopen. I also added the self-study tag and would recommend that you read its tag wiki. Here is a first hint to get you started: you need to get the expectation of $\hat{U}$. How can you calculate it, given that the $X_i$ are IID? – Stephan Kolassa Jan 4 '18 at 18:43
• the bias given in the example you cite $b=-U/(n+1)$ is wrong. There must be a couple typing mistakes in your book. – DeltaIV Jan 5 '18 at 10:20

The trickiest part is getting $b$ in terms of $U$ and $n$ so that we can express it as a function of parameters, rather than the random variables $\{{X_i}\}^{n}_{i=2})$. It helps to observe that expectation is linear. That is, $E(X_{1}+X_{2})=E(X_1)+E(X_2)$. Try applying this rule to the equation for $b$:
$b=E(\frac{X_2+X_3+⋯+X_n}{n+10})−U$