Sampling distribution of a statistic

  • #1
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Looking at another textbook problem, hope someone can let me know if I'm on the right track:

Let $X_1, X_2, ... X_{25}$ be a random sample from some distribution and let $W = T(X_1, X_2, ... X_{25})$ be a statistic. Suppose the sampling distribution of W has a pdf given by $f(x) = \frac{2}{x^2}, 1 < x < 2$ . Find the probability that W < 1.5

So from what I understand, we're sampling 25 numbers from a population and then getting some statistic, like a mean or a median from those 25 numbers. If you sample enough times, you get $f(x) = \frac{2}{x^2}, 1 < x < 2$ as a sampling distribution for that statistic, between 1 and 2. I'm pretty sure you need to take a definite integral here with 1.5 as an upper limit, but what's the lower limit? 1? 0? $-\infty$ ? Also, do you integrate with respect to x or some other variable? Or maybe I'm way off. Hope someone can help, thanks!
 

Answers and Replies

  • #2
Looking at another textbook problem, hope someone can let me know if I'm on the right track:

Let $X_1, X_2, ... X_{25}$ be a random sample from some distribution and let $W = T(X_1, X_2, ... X_{25})$ be a statistic. Suppose the sampling distribution of W has a pdf given by $f(x) = \frac{2}{x^2}, 1 < x < 2$ . Find the probability that W < 1.5

So from what I understand, we're sampling 25 numbers from a population and then getting some statistic, like a mean or a median from those 25 numbers. If you sample enough times, you get $f(x) = \frac{2}{x^2}, 1 < x < 2$ as a sampling distribution for that statistic, between 1 and 2. I'm pretty sure you need to take a definite integral here with 1.5 as an upper limit, but what's the lower limit? 1? 0? $-\infty$ ? Also, do you integrate with respect to x or some other variable? Or maybe I'm way off. Hope someone can help, thanks!

Hi das,

I interpret the specification of $f$ to mean that the density is 0 outside of the given interval.
What do you get if you integrate $f$ between $1$ and $2$?
 
  • #3
Hi! Thank you you've been very helpful.
Integrating between 1 and 2 gets the indefinite integral of $-\frac{2}{x}$. Between 2 and 1 this is (-1) - (-2) = 1. Which makes sense, but how do I take that info and figure out the probability that W < 1.5 ?

Thanks again
 
  • #4
Hi! Thank you you've been very helpful.
Integrating between 1 and 2 gets the indefinite integral of $-\frac{2}{x}$. Between 2 and 1 this is (-1) - (-2) = 1. Which makes sense, but how do I take that info and figure out the probability that W < 1.5 ?

Thanks again

It confirms that the density is indeed 0 outside the interval.
And it means that:
$$P(W<1.5) \underset{def}{=} \int_{-\infty}^{1.5} f_W(x) \,dx = \int_{1}^{1.5} f(x) \,dx$$
 
  • #5
But if the density is 1 between 1 and 2, doesn't that mean that it should be 0 everywhere else? And that's not true--If I integrate with different limits, say between 1 and 3, I get $\frac{4}{3}$. If the density were 0 outside this interval, doesn't that mean I should still get 1 between 1 and 3?
 
  • #6
But if the density is 1 between 1 and 2, doesn't that mean that it should be 0 everywhere else? And that's not true--If I integrate with different limits, say between 1 and 3, I get $\frac{4}{3}$. If the density were 0 outside this interval, doesn't that mean I should still get 1 between 1 and 3?

You are correct that the density is 0 elsewhere. So if you did something like: \(\displaystyle \int_{1}^{3}f(x) \,dx\) you would have to split it up into two integrals since $f(x)$ is not the same over that region.

\(\displaystyle \int_{1}^{3}f(x) \, dx = \int_{1}^{2} \frac{2}{x^2} \, dx+\int_{2}^{3}0 \, dx\)
 
  • #7
Ah ok thank you. Guess there's no reason you'd ever need to complicate your life by doing that.
 

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