MHB Sampling distribution of a statistic

AI Thread Summary
The discussion focuses on finding the probability that a statistic W, derived from a random sample of 25 numbers, is less than 1.5, given its probability density function (pdf) f(x) = 2/x^2 for 1 < x < 2. The user correctly identifies that the integral must be taken from 1 to 1.5 to compute this probability, confirming that the density is zero outside the specified interval. Clarifications are made regarding the integration limits and the necessity to split integrals when evaluating probabilities over ranges that include zero density. The conversation concludes with an understanding that integrating outside the defined limits is unnecessary and can complicate the calculation.
das1
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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!
 
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das said:
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 $2$?
 
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
 
das said:
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:
$$P(W<1.5) \underset{def}{=} \int_{-\infty}^{1.5} f_W(x) \,dx = \int_{1}^{1.5} f(x) \,dx$$
 
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?
 
das said:
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: $$\int_{1}^{3}f(x) \,dx$$ you would have to split it up into two integrals since $f(x)$ is not the same over that region.

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