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I'm trying to deduce from samples of observations from two finite sets of random variables $X_{1}, ..., X_{n}$ and $Y_{1}, ..., Y_{m}$ that the expected values of the average of those random variables are not the same. Let's denote $\frac{1}{n}\sum\limits_{i=1}^{n}{X_{i}}$ and $\frac{1}{m}\sum\limits_{i=1}^{m}{Y_{i}}$, $\bar{X}$ and $\bar{Y}$ respectively. So I'm trying to deduce from observations $x_{1}, ..., x_{n}$ and $y_{1}, ..., y_{m}$ that $E[\bar{X}] \neq E[\bar{Y}]$.

So my null hypothesis, $H_{0}$, would be $E[\bar{X}] = E[\bar{Y}]$ and the alternative one, $H_{1}$, would be $E[\bar{X}] \neq E[\bar{Y}]$. I thought of using Hoeffding's inequality this way:

If my random variables are all bounded and between $0$ and $1$ (which I have in practice), we then have: $P(|\bar{X}-\bar{Y}-(E[\bar{X}]-E[\bar{Y}])|\geq \varepsilon) \leq e^{-\frac{2\varepsilon^{2}}{\frac{1}{n}+\frac{1}{m}}}$

Under the null hypothesis we have then : $P(|\bar{X}-\bar{Y})|\geq \varepsilon) \leq 2e^{-\frac{2\varepsilon^{2}}{\frac{1}{n}+\frac{1}{m}}}$, we can fix the type $I$ error rate to a maximum of $\delta$ if we take $\varepsilon_{\delta} = \sqrt{\frac{1}{2(\frac{1}{n}+\frac{1}{m})}ln(\frac{2}{\delta})}$. Now, suppose we get observations such as $|\bar{X}-\bar{Y}| < \varepsilon_{\delta}$, there's a probability of $P(|\bar{X}-\bar{Y}| < \varepsilon_{\delta}) = 1 - P(|\bar{X}-\bar{Y}| \geq \varepsilon_{\delta})$ under $H_{0}$, so our type $II$ error is at most $1 - \delta$.

I hope you can help me understand if what I'm doing is correct or not. Thank you in advance.

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  • $\begingroup$ When $n$ and $m$ are $6$ or larger (and, of course, $0\lt\delta\le 1$), $\varepsilon_\delta$ exceeds $1:$ wouldn't that make your test useless? $\endgroup$
    – whuber
    Mar 27 at 16:52
  • $\begingroup$ @whuber thank you for your comment. Can you clarify useless in what sense please? When $\varepsilon_{\delta} \geq 1$ what we have is the probability that the difference in the means is greater in absolute value than some number greater than one is less than the $\delta$ right? Is the test useless in the sense that minor but significant changes in the means won't be taken into account? And can you tell me please if the reasoning in general is correct and just the test itself is useless or is it also an issue in reasoning? (Mainly talking about the type $I$ and $II$ errors). Thank you. $\endgroup$
    – Daviiid
    Mar 27 at 21:09
  • $\begingroup$ Useless in that it accomplishes nothing: since this inequality relies on assuming the distribution is supported on the interval $[0,1],$ that assumption alone implies the difference in means is no greater than $1.$ You don't need any data to figure that out! $\endgroup$
    – whuber
    Mar 28 at 1:06
  • $\begingroup$ @whuber you're right thank you! $\endgroup$
    – Daviiid
    Mar 28 at 10:06

1 Answer 1

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Your $\epsilon$ increases with $m$ and $n$ for fixed $\delta$, so the test becomes less powerful with more data. There must either be an error in the calculation or in the idea.

Starting with $$P(|\bar X-\bar Y|>\epsilon)\leq 2\exp\left( \frac{-2\epsilon^2}{\frac{1}{n}+\frac{1}{m}}\right)$$ we're going to write $\alpha$ for the right-hand side, because we'll need $\delta$ for more standard purposes late. I'm also going to write $\tilde n$ for $1/(1/n)+(1/m)$ to cut down on the opportunities for getting the reciprocals wrong (and cut down on the typesetting). We want to guarantee the LHS is less than $\delta$, so we want $$\alpha \leq 2\exp\left( -2\tilde n\epsilon^2\right)$$ Rearranging: $$\frac{\alpha}{2}\leq exp\left( -2\tilde n\epsilon^2\right)$$ $$\log \frac{\alpha}{2}\leq -2\tilde n\epsilon^2$$ $$\frac{1}{2\tilde n}\log \frac{\alpha}{2}\leq -\epsilon^2$$ $$-\frac{1}{2\tilde n}\log \frac{\alpha}{2}\geq \epsilon^2$$ $$\sqrt{-\frac{1}{2\tilde n}\log \frac{\alpha}{2}}\geq \epsilon$$ $$\sqrt{\frac{1}{2\tilde n}\log \frac{2}{\alpha}}\geq \epsilon$$ which is like your bound except it decreases with increasing sample size.

Now, the other thing that's wrong is that you haven't specified a distance of departure from the null in getting your power -- the power must decrease to $\delta$ as $E[X]-E[Y]\to 0$, so you need to specify what $|E[X]-E[Y]|$ is under the alternative for which you want the specified power. Without loss of generality, suppose $X$ has the larger mean and write $\delta=E[X]-E[Y]$. Also write $Z$ for $X-\delta$, so that the means of $Z$ and $Y$ are the same. Now, $\bar Z-\bar Y$ is controlled by Hoeffding's Theorem, since $Z$ is still bounded, and if $$P[|\bar X-\bar Y| >\delta/2]\leq \alpha$$ under the null and $$P[|\bar Z-\bar Y| >\delta/2]\leq \alpha$$ under the alternative, then you have a useful test, since at most one of $|\bar X-\bar Y|>\delta/2$ and $|\bar Z-\bar Y|>\delta/2$ can be true

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  • $\begingroup$ Thank you for your answer. It seems I'm wrong both in the calculations and the idea. I think I understood clearly your answer but there are some obscure points I'd really appreciate you clarifying. So here $\delta$ is equal to $2 \varepsilon$ right? Does the $1/2$ have a particular meaning or can we take whatever number it is? And with that we have a rate of $\alpha$ for both type $I$ and $II$ errors right? Finally can you clarify why at most one of $|\bar{X}-\bar{Y}|>\delta/2$ or $|\bar{Z}-\bar{Y}|>\delta/2$ please? Thank you again. $\endgroup$
    – Daviiid
    Mar 28 at 0:16
  • $\begingroup$ Because $\bar X$ and $\bar Z$ differ by $\delta$, $\bar Y$ can't be closer than $\delta/2$ to both of them at once. The point of using $\delta/2$ is exactly to make this happen, though you could use two other numbers that add up to $\delta$. $\endgroup$ Mar 28 at 2:54
  • $\begingroup$ I see, thank you again that's really helpful. $\endgroup$
    – Daviiid
    Mar 28 at 10:07
  • $\begingroup$ I'd like to add (I couldn't edit the last comment I'm sorry), that the test you came up with is really interesting since it lets us control both types of errors with the same rate $\alpha$. Generally I find it's $\alpha$ and $1-\alpha$ so it leads us to a trade-off dilemma. $\endgroup$
    – Daviiid
    Mar 28 at 10:25

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