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We have $X_i \sim ^{iid} N(\mu,\sigma^2)$with known mean $\mu$, and unknown $\sigma^2$. Let's suppose we're given $s^2$ (sample variance) and $n$ (sample size).

We know $\frac{\bar X-\mu}{S/\sqrt{n-1}} \sim t(n-1)$.

  • Is there a way to compute $P(\bar X>c)$?

Now, I've seen several times the following reasoning:

$$P(\bar X>c)=P\left(\frac{\bar X-\mu}{S/\sqrt{n-1}}>\frac{c-\mu}{s/\sqrt{n-1}}\right)$$

However, I find this odd, instead I think we should have

$$P(\bar X>c)=P\left(\frac{\bar X-\mu}{S/\sqrt{n-1}}>\frac{c-\mu}{S/\sqrt{n-1}}\right)$$

because we're supposed to divide by the same expression on both sides of the inequality.

(Note: By $s$ I mean the r.v $S$ evaluated at a given sample)

The motivation for this question is the following: Imagine I wanted to test the null $\mu \leq \mu_0$. If we knew the population variance, then we would expect to reject the null if $\bar X > c'>\mu_0$. So,

$$P(\bar X>c')=P\left(\frac{\bar X-\mu_0}{\sigma/\sqrt{n-1}}>\frac{c'-\mu_0}{\sigma/\sqrt{n-1}}\right)$$

and we would chose $\frac{c'-\mu_0}{\sigma/\sqrt{n-1}}:= q_{1-\alpha}$, the $1-\alpha$ quantile for $N(0,1)$.

However, this type of reasoning cannot be done for the case at the begining of this question. We can prove that a similar rejection interval does indeed come out from using the Likelihood Ratio Test(LRT). I was wondering if there was a way to reach the same rejection interval without having to use the LRT...

Any help would be appreciated.

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    $\begingroup$ What is the difference between $S$ and $s$? Are you sure it isn't a typo? $\endgroup$
    – dlnB
    Commented Mar 14, 2019 at 21:54
  • $\begingroup$ @dlnB The difference is that $s$ is $S$ computed for a given sample, i.e. it's a constant. $S$ is not. $\endgroup$ Commented Mar 14, 2019 at 21:57
  • $\begingroup$ $P(\bar{X}_n>c) =P(\frac{\bar{X}_n-\mu}{s/\sqrt{n-1}}>\frac{c-\mu}{s/\sqrt{n-1}})$ is correct logic. Multiplying both sides by a positive constant or subtracting a constant from both sides preserves the inequality. $\endgroup$
    – dlnB
    Commented Mar 14, 2019 at 22:19

1 Answer 1

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$X_1,...,X_n$ are iid $N(\mu, \sigma^2)$. The first step is to derive the distribution of $\bar{X}_n= \sum_{i=1}^n \frac{1}{n}X_i$. First notice that $\frac{1}{n}X_i \sim N(\frac{\mu}{n}, \frac{\sigma^2}{n^2})$ for $i=1,...,n.$ Then $\bar{X}_n$ is simply the sum of these iid variables $\frac{1}{n}X_i$, so $\bar{X}_n \sim nN(\frac{\mu}{n} ,\frac{\sigma^2}{n^2})=N(\mu ,\frac{\sigma^2}{n}).$ It follows that $$P(\bar{X}_n > c) = \int_{c}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2/n}} \exp\{-\frac{(z-\mu)^2}{2\sigma^2/n}\}dz. $$

This is one expression for the exact probability you describe, but requires $\sigma^2$ for calculation.

Following the $t$ approach you describe, $$P(\bar{X}_n>c) =P(\frac{\bar{X}_n-\mu}{s/\sqrt{n-1}}>\frac{c-\mu}{s/\sqrt{n-1}})=P(t(n-1) > \frac{c-\mu}{s/\sqrt{n-1}}),$$

which can simply be found by taking $1-F(\frac{c-\mu}{s/\sqrt{n-1}})$, where $F(\cdot)$ is the cdf for the $t(n-1)$ distribution.

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  • $\begingroup$ thanks for the answer, however, I don't think it's correct... we need to use $S$ and not $s$ for the pivot to have the $t(n-1)$ distribution. $\endgroup$ Commented Mar 14, 2019 at 22:30
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    $\begingroup$ I still don't understand what the two different notations mean. You write $s^2$ is sample variance, but use $S$ in your formula for $t(n-1)$. The "s" or "S" used in the $t$ formula is the estimated standard deviation from the sample. $\endgroup$
    – dlnB
    Commented Mar 14, 2019 at 22:33
  • $\begingroup$ It's the same difference as the one we have between $X_i$ and $x_i$. Hum... I'm not going to be completely correct, but think of $X_i$ as a random variable and $x_i$ as an observation of the variable $X_i$. $x_i$ is a constant. $\endgroup$ Commented Mar 14, 2019 at 22:37
  • $\begingroup$ I see your confusion. $\bar{X}$ and $S$ are indeed random variables, so let $\bar{X}_n$ and $s$ denote realized values of those random variables, calculated from your sample $X_1,...,X_n$. A sample $X_1,...,X_n$ gives us realized values $\bar{X}_n$ and $s$ and therefore a realization of $\frac{\bar{X}_n-\mu}{s/\sqrt{n-1}}$, which is a realized value from a $t(n-1)$ distribution. Therefore $P(\frac{\bar{X}_n-\mu}{s/\sqrt{n-1}}>\frac{c-\mu}{s/\sqrt{n-1}})=P(t(n-1) > \frac{c-\mu}{s/\sqrt{n-1}}).$ $\endgroup$
    – dlnB
    Commented Mar 14, 2019 at 22:48
  • $\begingroup$ The problem is that we have only $\frac{\bar X-\mu}{S/\sqrt{n-1}}\sim t(n-1)$. We do NOT know the distribution of $\frac{\bar X-\mu}{s/\sqrt{n-1}}$ (because we do not know the population variance) $\endgroup$ Commented Mar 14, 2019 at 23:00

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