6
$\begingroup$

I was not able to find this question here, so I am going to ask this:

  • What is the difference between $\mathbb{E}(\bar{X})$ (expected value of $X$ bar) and the actual $\bar{X}$? I am very confused about these two concepts. How come $\bar{X}$ is one of the estimators of normal distribution (the other one being $S^2$) and then what is the point of $\mathbb{E}(\bar{X})$?

  • Another question that I have is, what is the relationship between $\bar{X}$ and $E(\widehat{\mu})$ ? I understand that mu is a true mean, but then what is $E(\widehat{\mu})$?

    I would appreciate any explanation on these concepts and since I'm only a beginner in math stats, I am struggling with notation, so any simplified explanation will be much appreciated!

    And then, I just discovered that there is also an expected value of $\sigma^2$ which, why lie, completely blew my mind! So I suspect I'm struggling with the whole concept of expected value and how it relates to the the population, sample distribution, and sampling distribution, so I would definitely appreciate an explanation of expected value of $\bar{X}$ in light of this.

$\endgroup$
  • $\begingroup$ @user777 thank you,it was very helpful, it's a great opportunity for me to finally start learning Latex! $\endgroup$ – Jen Sep 12 '14 at 17:35
  • $\begingroup$ yeah, I'm able to see that, that's how I was able to edit all the other $\bar{X}$ 's $\endgroup$ – Jen Sep 12 '14 at 17:39
8
$\begingroup$

First of all, $\bar{X}$ is not an estimator of the Normal distribution. The true mean $\mu$ (as well as the true variance $\sigma^2$) is a parameter of the Normal distribution. $\bar{X}$ is an estimator for $\mu$ and this distinction is extremely important.

It sounds like that your confusion stems from not understanding that estimators themselves have distributions.

Suppose you have a true mean $\mu$ and you try to estimate what that true $\mu$ is. You do this by getting data through an experiment and compute $\bar{X}$. But $\bar{X}$ may not necessarily equal $\mu$; in other words, there is a variance to $\bar{X}$. So although the true $\mu$ may be $10$, your $\bar{X}$ may be $10.1$, or $9.7$ or some other value. So you should think of $\mathbb{E}(\bar{X})$ as the mean of the estimator of the mean. So since $\mathbb{E}(\bar{X}) = \mu$, we know that our realization of the sample mean (the $\bar{X}$) will be drawn from a distribution around $\mu$.

It would be fantastic to plug in $\mathbb{E}(\bar{X})$ to the Normal distribution since it equals $\mu$, but we don't know what $\mathbb{E}(\bar{X})$ is since the data can only tell us the realization of $\bar{X}$.

As to your second question, sometimes people write $\hat{\mu}$ is mean $\bar{X}$. The hat denotes that $\hat{\mu}$ is an estimator for $\mu$, which is what $\bar{X}$ is.

Like above, the $\mathbb{E}(\hat{\sigma}^2)$ (notice the hat) is the expected value of the estimator $\hat{\sigma}^2$, which is an estimator for the parameter $\sigma^2$ as you wrote above.

$\endgroup$
  • 1
    $\begingroup$ $E(\bar{x}) = \mu$: the mean of the mean is the mean! $\endgroup$ – Aaron left Stack Overflow Sep 12 '14 at 18:42
  • $\begingroup$ @BlueMarker , thank you so much, it was extremely helpful and clear!I only have one confusion at this point, and I'm sorry if it sounds too silly, but what's the actual point of finding those distributions and expected values of $\hat{\sigma}^2$ or $\mathbb{E}(\bar{X})$ ? In other words, what is the practical purpose of it? Why would you need them? In order to estimate the true variance and mean or is there some formal reason and explanation of this necessity? $\endgroup$ – Jen Sep 12 '14 at 18:59
  • $\begingroup$ @Ana So in any experiment, we'll probably never know the true $\mu$ (that's why statistics exists, after all), but it is important to know $\mathbb{E}(\bar{X})=\mu$ since that tells us that the estimator $\mathbb{E}(\bar{X})$ has one good property. It tells us that even though we don't know if the realized $\bar{X}$ will be equal on $\mu$, if I had to guess at what it will be equal to, I would guess that it's equal to $\mu$. So if we were to have a generic parameter $\theta$ and an estimator $\hat\theta$, it might be that $\mathbb{E}(\hat\theta)=\theta+1000$. $\endgroup$ – Blue Marker Sep 12 '14 at 19:11
  • 1
    $\begingroup$ @BlueMarker Thank you!That makes sense. But then I have another question, when I set up a null hypothesis, there is also this other parameter that confuses me even more, $\tilde{\mu}$ . Such as in this relationship, $$\Pr\left (\frac{\bar{X}-{\tilde{\mu}}}{\sqrt{\sigma_0/{\sqrt{n}}}} < -Z_{\alpha/2} \right ) + \Pr\left (\frac{\bar{X}-{\tilde{\mu}}}{\sqrt{\sigma_0/{\sqrt{n}}}} > Z_{\alpha/2} \right ) = \alpha$$ !!Then what is the difference between $\bar{X}$ and $\tilde{\mu}$ in this case? Side Note:[I can't believe I just wrote this whole thing myself in LateX!!Life-changing!] $\endgroup$ – Jen Sep 12 '14 at 20:49
  • $\begingroup$ and $H_0 : \mu_0 = \tilde{\mu}$ $\endgroup$ – Jen Sep 12 '14 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.