Biased Estimates: û : (X1 + 2X2 + 3X3)/6 & ü : (X1 + 4X2 + X3)/6

[Alex_Of_Darkness]

Biased Estimates: û : (X1 + 2X2 + 3X3)/6 & ü : (X1 + 4X2 + X3)/6

Hi everyone, hopefully you'll be able to help me with this problem that is giving me headaches.

X1, X2, X3 are three random variables from a population with a mean "u" and a variance "σ2".

There is two estimators : û : (X1 + 2X2 + 3X3)/6 & ü : (X1 + 4X2 + X3)/6

I need to find if they are biased or not, if they are I need to calculate its bias.

I know that if u = û and u = ü there will be no bias but since we don't know anything about u I have no clue how to solve that. There is also 3 other subquestions but I guess that if I understand how to solve this one, I might be able to solve the other three by myself.

Thanks a lot in advance everyone!

 

[tkhunny]

Hi everyone, hopefully you'll be able to help me with this problem that is giving me headaches.

X1, X2, X3 are three random variables from a population with a mean "u" and a variance "σ2".

There is two estimators : û : (X1 + 2X2 + 3X3)/6 & ü : (X1 + 4X2 + X3)/6

I need to find if they are biased or not, if they are I need to calculate its bias.

I know that if u = û and u = ü there will be no bias but since we don't know anything about u I have no clue how to solve that. There is also 3 other subquestions but I guess that if I understand how to solve this one, I might be able to solve the other three by myself.

Thanks a lot in advance everyone!

What do you get for the Expected Value of your estimators?

Can we assume \(\displaystyle E[X_{1}] = \mu\)?

 

[Alex_Of_Darkness]

What do you get for the Expected Value of your estimators?

Can we assume \(\displaystyle E[X_{1}] = \mu\)?

I'm not sure how I could calculate expected value with no information on X1, X2 and X3?

And why could we assume that \(\displaystyle E[X_{1}] = \mu\)?

 

[ksdhart2]

I'm not sure how I could calculate expected value with no information on X1, X2 and X3?

You have lots of information about X₁, X₂, and X₃. It's all in the problem text.

And why could we assume that \(\displaystyle E[X_{1}] = \mu\)?

Because the problem tells you as much. When in doubt, return to the definitions of the terms involved.

X1, X2, X3 are three random variables from a population with a mean \(\displaystyle \mu\) and a variance \(\displaystyle \sigma^2\)

So you have three random variables. Well, what's a random variablehttps://www.mathsisfun.com/data/random-variables.html? I've linked a page from the website MathIsFun if you need a refresher. More specifically, what is the expected value of a random variable? I mean just in general, the concept, not any specific random variable.

As an example, suppose you toss 3 coins. Let X be the number of heads observed. How would we calculate the expected value of X? Let's use the formula: \(\displaystyle \displaystyle E(X) = \sum_{x \in X} x \cdot P(X = x)\). We know P(X = 3) = 1/8; P(X = 2) = 3/8; P(X = 1) = 3/8; and P(X = 0) = 1/8. That means \(\displaystyle E(X) = 3 \cdot \dfrac{1}{8} + 2 \cdot \dfrac{3}{8} + 1 \cdot \dfrac{3}{8} + 0 \cdot \dfrac{1}{8} = \dfrac{3}{8} + \dfrac{6}{8} + \dfrac{3}{8} = \dfrac{12}{8} = \dfrac{3}{2}\)

That process kinda looks like finding an average. Indeed, another term for "expected value" of a random variable is "weighted average." And what's another term for "average?"

 

[Alex_Of_Darkness]

​Error

 

[Alex_Of_Darkness]

So sorry for the double post, I think the forum was having some issues so I was not able to post what I wanted.


Another term would be the mean which is equal to u so by finding the expected value I will also be able to find if it is equal to u.


If I understand correctly would my X1, X2 and X3 like the three coins you talked about in your exemple? Would it be something like this? :
But what do I put for the amount before each probability? In your example I understand that it's the amount of time the event in question happens but here would in be only one?
Probability for X1 would be 1/6, 2/6 for X2 and 3/6 for X3
1x1/6 + 1x2/6 + 1x3/6 = that would be one which I'm pretty sure is wrong.

Also, now since you showed me that u=X1 I guess that for an estimator not to be biased I would have to find X1 as its value?

Thanks a lot for taking the time by the way, you cannot even imagine how grateful I am because like I said this problem has a lots of subquestion and I'm sure that if I get through this first part I'll be fine for the rest.

 

[Alex_Of_Darkness]

This problem has been solved, thanks to all for your help