I'm trying to replicate some math a professor did related to Twitter sentiment analysis. Basically, there is a sentiment dictionary, called ANEW, that contains a mean and standard deviation for 3 different types of scores: valence, arousal, and dominance. The link above is to the page with a full description (with the germane math being 3/4 of the way down). The pertinent section is also available here, in this image:

enter image description here

I understand what he did with calculating the overall standard deviation. However, I do not understand what he did with calculating p(sub-i) in the second set of equations. It looks like he's taking the product of a series, but what series? If I plug in a value for i, then am I not just getting p for the standard deviation of the valence value at position i?

I have tried to reproduce his example in the final paragraph at least 4 or 5 different ways, but I never end up with the same values. Given the values presented in the final paragraph of the image, can somebody walk me through how to calculate the overall weighted average of the valence for the tweet presented?

  • $\begingroup$ I think that's supposed to be a plain old $\pi$, not a product. $\endgroup$ – tabstop Jan 28 '14 at 4:53
  • $\begingroup$ @tabstop I tried that... but, for instance, for the term "health," as shown in the last paragraph, 1/(2*pi*1.88*1.88) = 0.045, which doesn't align with his calculations. $\endgroup$ – Clay Jan 28 '14 at 4:56
  • $\begingroup$ Checking with wikipedia (I know, I know), the maximum value of the normal distribution is the square root of the listed value (i.e. $1/(\sigma\sqrt{2\pi})$) and that appears to give the values listed. $\endgroup$ – tabstop Jan 28 '14 at 5:06
  • $\begingroup$ @tabstop Aha! Thank you! The formula is incorrect. I'm happy to mark the answer as correct if you submit it! I'll inform the professor. Thanks! $\endgroup$ – Clay Jan 28 '14 at 5:17

The peak of a normal distribution is $1/(\sigma\sqrt{2\pi})$, which is the square root of the listed value.

  • $\begingroup$ Thanks again. I informed the professor and he updated his page. $\endgroup$ – Clay Jan 28 '14 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.