# Finding the expectation of a constant bias

• Apr 10th 2009, 09:26 AM
mongrel73
Finding the estimate of a constant bias on measurements
Someone is making four measurements:
$y_1$, $y_2$ of $\lambda$, $\mu$ respectively, $y_3$ and $y_4$ both of $\lambda + \mu$.
The measurements are subjected to independent, normally distributed random errors with known variance ${\sigma}^2$.
He suspects that there is a constant bias $\beta$
How do you show that this bias is estimated as $y_1+y_2-\tfrac{1}{2}(y_3+y_4)$ ?
• Apr 10th 2009, 12:29 PM
CaptainBlack
Quote:

Originally Posted by mongrel73
Someone is making four measurements:
$y_1$, $y_2$ of $\lambda$, $\mu$ respectively, $y_3$ and $y_4$ both of $\lambda + \mu$.
The measurements are subjected to independent, normally distributed random errors with known variance ${\sigma}^2$.
He suspects that there is a constant bias $\beta$
How do you show that this bias is estimated as $y_1+y_2-\tfrac{1}{2}(y_3+y_4)$ ?

Suppose:

$y_1=\lambda+\beta+\varepsilon_1$

$y_2=\mu+\beta+\varepsilon_2$

$y_3=\lambda+ \mu +\beta+\varepsilon_3$

$y_3=\lambda+ \mu +\beta+\varepsilon_4$

where $\varepsilon_1,\ \varepsilon_2,\ \varepsilon_3,\ \varepsilon_4$ are normally distributed with zero mean.

Now write:

$\theta= y_1+y_2+(y_3+y_4)/2$

Now find $E(\theta)$ and show that this is $\beta$

CB
• Apr 10th 2009, 05:00 PM
mongrel73
Quote:

write:

$\theta= y_1+y_2-(y_3+y_4)/2$

Now find $E(\theta)$ and show that this is $\beta$
Doesn't that just show that the estimate is unbiased?
I mean, if you write $\theta=y_1+y_2-y_3$, then $E(\theta)=\beta$ as well, so that can't be sufficient.
How do you show that the one given in the question is the best estimate to use?
• Apr 11th 2009, 12:04 AM
CaptainBlack
Quote:

Originally Posted by mongrel73
Doesn't that just show that the estimate is unbiased?
I mean, if you write $\theta=y_1+y_2-y_3$, then $E(\theta)=\beta$ as well, so that can't be sufficient.
How do you show that the one given in the question is the best estimate to use?

1. $E(\theta)=\beta$ is the condition that \theta is an (unbiased (Evilgrin)) estimator of the bias.

2. The question did not ask that you show it is the best estimate, nor specify in what sense it should be "best".

CB