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$$u (\text{mean}) = \frac{X_1} 5 + \frac 4 {(5N-1)} \cdot (X_2 +X_3 + \cdots + X_N)$$

I'm supposed to find the bias and variance of this estimator, but not sure how to do this. The sample is independent and normally distributed. This is a review problem set and we didn't cover this in class, so I'm a bit rusty. Would appreciate guidance.

What I have so far:

$$\frac{5X_1}{5N} - \frac{NX_1}{5N} = \frac{X_1(5-N)}{5N}$$

$$ \frac{X_2 +X_3 + \cdots + X_N} N - \frac 4 {5N-1} \cdot (X_2 +X_3 + \cdots + X_N) = \frac 1 5 \cdot \frac{X_2 +X_3 + \cdots + X_N} N$$

So bias: $ (X_1(5-N)/5N) + (1/5) (X_2 +X_3 + \cdots + X_N)/N$ right?

I totally forgot how to find variance, would appreciate guidance on this. What I have so far on variance:

$$\text{Var} = \frac 1 N \left(\sum_i X_i^2 - \left[N \cdot \frac 4{5N-1} \cdot (X_2 +X_3 + \cdots + X_N)\right]^2\right)$$

that's all I have so far. Do I need to simplify further? Also that's all I need to find MSE right?

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    $\begingroup$ What is the to value of interest which hast to be estimated? What is the estimator? Are $X_1, ..., X_N$ independent and identically distributed? $\endgroup$
    – Obriareos
    Commented Feb 12, 2017 at 21:10
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    $\begingroup$ It is impossible to say what the bias is without knowing what is being estimated or without knowing anything about the probability distributions involved. $\endgroup$ Commented Feb 12, 2017 at 21:59
  • $\begingroup$ Sorry, it's basically using the equation above as a bias for the mean. $\endgroup$ Commented Feb 12, 2017 at 23:03
  • $\begingroup$ Is there any chance that $5(N-1)$ was intended, rather than $5N-1$? $\endgroup$ Commented Feb 12, 2017 at 23:26
  • $\begingroup$ Yeah, thanks for clarifying $\endgroup$ Commented Feb 13, 2017 at 0:16

2 Answers 2

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For a computation of the bias you should to tell us the parametric distribution of the (iid?) random variables $X_i$. (I post this as an answer because my reputation is not sufficient to post it as a comment).

EDIT: Ok, then let $\mu$ and $\sigma$ be the parameters of the normal distributed iid random variables $X_i$. According to the comments above the corrected estimate is:

$$U(X_1,...,X_N)=\frac{X_1}{5}+\frac{4}{5(N-1)}(X_2+...+X_N),$$

with a slight modification of the denominator $(5N-1)\to5(N-1)$ (which make sense). Then, the expectation value is

$$\text{E}[U(X_1,...,X_N)]=\frac{\mu}{5}+\frac{4}{5(N-1)}(N-1)\,\mu=\mu.$$

That is, the estimator is unbiased since $\text{E}[U-\mu]=0$.

Similar to the Variance:

$$\text{Var}[U]=\left(\frac{1}{5}\right)^2\sigma^2+\left(\frac{4}{5(N-1)}\right)^2(N-1)\,\sigma^2=\frac{N+15}{25(N-1)}\,\sigma^2.$$

For large $N\to\infty$ the variance approaches to

$$\text{Var}[U]\to\left(\frac{\sigma}{5}\right)^2$$

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  • $\begingroup$ Right. A Comment, not an Answer. But a good comment. (+1) towards being able to comment. $\endgroup$
    – BruceET
    Commented Feb 12, 2017 at 22:43
  • $\begingroup$ Each X is an observation in a sample that's independent and normally distributed $\endgroup$ Commented Feb 12, 2017 at 23:08
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The fundamental properties you need are as follows:

$$E(aX + bY) = aE(X) + bE(Y).$$ This extends to more than two random variables. Do your $X_i$ all have the same mean $\mu$? I'll assume so.

$$Var(aX + bY) = a^2Var(X) + b^2Var(Y),$$ provided $X$ and $Y$ are independent. So as noted by @kaffeeauf, you need to specify that the $X_i$ are independent.

Now, for your random variable $$U = \frac{X_1} 5 + \frac 4 {5n-1} \cdot (X_2 +X_3 + \cdots + X_n)$$

To start, let $U_1 = \frac15 X_1.$ Then $E(U_1) = \frac 15 E(X_1) = \frac 1 5 \mu.$

Now let $U_2 = \frac{4}{5n-1}(X_2 \dots X_n).$ Then $E(U_2) = \frac{4(n-1)}{5n-1}E(X_i) = \frac{4(n-1)}{5n-1}\mu.$

Can you take it from there to find $E(U) = E(U_1 + U_2)?$

As for unbiasedness, that only makes sense if the $X_i$ are iid with $E(X_i) = \mu$ and you are considering $U$ as an estimator of $\mu.$ Then the bias of $U$ is $b_U = E(U) - \mu.$ (See the Comments by @MichaelHardy and @Obiareos.) The question keeps mutating, I hope this matches the current version. Of course, the 'usual' estimator of $\mu$ would be $\bar X,$ which is unbiased because $E(\bar X) = \mu.$

Then in the last part on variances, I suppose you will find that $Var(cU) > Var(\bar X) = \sigma^2/n,$ where $\sigma^2 = Var(\bar X),$ where $c$ is chosen so $E(cU) = \mu.$

Note: Everything considered, I think your review will go better if you review the topics of interest before trying to work exercises on those topics.

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  • $\begingroup$ So is bias just $ \frac{4(n-1)}{5n-1}\mu. + 4/5mu$ ? Thanks for helping out- how do I deal with variance now? $\endgroup$ Commented Feb 13, 2017 at 0:44
  • $\begingroup$ Assuming independence, similar approach, but note squared coefficients in 2nd displayed eqn. $\endgroup$
    – BruceET
    Commented Feb 13, 2017 at 4:06
  • $\begingroup$ Yeah but what exactly do I do? Can you please go a step further I'd really appreciate it $\endgroup$ Commented Feb 13, 2017 at 16:52
  • $\begingroup$ To start, $U_1 = \frac15 X_1.$ Then $V(U_1) = (\frac 15)^2 V(X_1) = \frac {1} {25} \sigma^2.$ Similarly for my $U_2.$ Messier algebra, but essentially the same process. $\endgroup$
    – BruceET
    Commented Feb 13, 2017 at 17:42

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