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Let $X_{1},\cdots,X_{n}$ be random samples which has normal distribution $N(\mu,\sigma^{2})$.

When $\mu$ and $\sigma^{2}$ are unknown, I want to find UMVU estimator for $\frac{\mu}{\sigma}$.

I know that $(\sum_{i=1}^{n}X_{i},\sum_{i=1}^{n}X_{i}^{2})$ is complete sufficient statistic for $\left(\mu,\sigma^{2}\right)$.

Let $\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$, $S^{2}=\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}$. Then $E(\bar{X})=\mu$ and $E(S^{2})=\sigma^{2}+\mu^{2}$.

So for $\frac{\mu}{\sigma}$, I guess as a estimator $$ Y^{2}=\frac{\bar{X}^{2}}{\frac{1}{n}\sum_{i=1}^{n}X_{i}^{2}-\bar{X}}. $$ But how can I compute $E(Y^{2})?$

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  • $\begingroup$ Not quite awake yet, but I am surprised to see $\bar X$ in the denominator, and to be told that it has a gamma distribution. $\endgroup$ – BruceET Dec 18 '16 at 18:11
  • $\begingroup$ @BruceET Why does that follow gamma distribution? I know what gamma distribution is and I studied about it. But I can't get any point about this. Actually the other person who repeated to my question told it to me about 'gamma distribution'. $\endgroup$ – kayak Dec 19 '16 at 1:36
  • $\begingroup$ I will not go into detail about specific gamma distributions. For properly standardized normal data $X_i$, you have that $\bar X$ is chi-squared, which is a member of the gamma family. Likewise, the sum of squared normals is also related to chi-squared. Not sure exactly what @Kwerty has in mind. Maybe he/she will elaborate. $\endgroup$ – BruceET Dec 19 '16 at 3:46
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Qwerty's answer is plain wrong, and BruceET's answer is incomplete. I will leave the computations to you but a reasonable guess for UMVUE of $\mu/\sigma$ would be $T=C\dfrac{\overline{X}}{S}$ where $C$ is a constant to be computed by you.

First note that UMVUE's are based on complete sufficient statistics (if they exist), and since $\overline{X}$ and $S^2$ are complete and sufficient, if you can find $C$ that makes $E(T)=\mu/\sigma$, then $T$ is going to be your UMVUE.

So how to compute $C$? Note, $E(T)=CE(\dfrac{\overline{X}}{S})=CE(\overline{X})E(\dfrac{1}{S})$ since for Normal distribution, $\overline{X}$ and $S$ are independent.

You know $E(\overline{X})=\mu$. Now here's a trick to compute $E(\dfrac{1}{S})$.

If $R^2\sim \chi^2_k$ then you can explicitly calculate the pdf of $\dfrac{1}{R}$, which is called inverted Gamma, and once you find the pdf of $\dfrac{1}{R}$ you can easily compute $E(\dfrac{1}{R})$ by working out an integral.

Note here that $nS^2/\sigma^2\sim \chi^2_{n-1}$. Hence you can find $E(1/S)$ and putting $E(T)=\mu/\sigma$ you can therefore find $C$.

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  • $\begingroup$ Ok I don’t know how that happened. Let me delete the unvoted answer lol. $\endgroup$ – Landon Carter Jun 24 '18 at 15:56
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Comment. It seems to me you may not be on a useful track. Maybe this will help.

Let $\bar X$ be the sample mean, $S^2$ be the sample variance, and $\tau = \mu/\sigma.$ Then by the method of moments it might be reasonable to look at $\hat \tau = \bar X/S$ as an estimate of $\tau$ based on sufficient statistics. One would not necessarily expect it to be an unbiased estimator (perhaps asymptotically unbiased), but might check to see how biased it is. Because $S$ is known to be slightly negatively biased for $\sigma$ it is not surprising that $\hat \tau$ is slightly positively biased for $\tau.$

I do not want to claim this is the path you should take, but I think it may make more sense than what you have suggested. (Your denominator does not seem feasible because your $S^2$ and $\bar X$ have different dimensionalities.)

With a quick simulation of a million samples of size $n = 10$ from $Norm(\mu = 100, \sigma=20)$ (so that $\tau = 5$), in R statistical software, we can investigate this idea. At the end of my simulation I have shown a modification of your estimator that may have promise.

m = 10^6;  mu = 100; sg = 20;  n = 10
x = rnorm(m*n, mu, sg);  DTA = matrix(x, nrow=m)              # each row a sample
a = rowMeans(DTA);  s = apply(DTA, 1, sd);  tau.hat = a/s     # m-vectors
mean(a);  sd(a);  mean(s);  sd(s)
## 99.99136  # exact is 100
## 6.313811  
## 19.45392  # as known, S slightly negatively biased for sg
## 4.640816
mean(tau.hat)
## 5.469827  # not surprisingly, somewhat positively biased for tau

# investigating your estimator
SS = rowMeans(DTA^2)            # your 'S^2'
your.est = a^2/(SS - a)
mean(your.est)
## 0.9745633                    # nowhere near tau
alt.est = sqrt(a^2/(SS - a^2))  # possible candidate
mean(alt.est)
## 5.765703                     # biased, but in the 'ballpark'
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  • $\begingroup$ You suggest $\bar{X}/S^2-\bar{X}^2$. Ah yes. I made mistake. I should put square there. But another one is : Is the condition 'complete and sufficient' is okay if I combinate two for a one parameter? $\endgroup$ – kayak Dec 19 '16 at 1:34
  • $\begingroup$ If I understand your question, Yes. As I recall, the idea is that the estimator should be a function of complete, sufficient statistics. If you want to use the 'alternate' estimator in my simulation, can you figure out the constant that 'unbiases' it?' I'm glad to have gotten you off of a wrong track, but further help should come from others. $\endgroup$ – BruceET Dec 19 '16 at 3:58

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