Wikipedia:Reference desk/Archives/Mathematics/2008 August 13

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August 13

Convolution

I know that the convolution with the Delta function has the property (I'm working in the frequency domain): $G(f)*\delta (f-a)=G(f-a)$

But what will $G(kf)*\delta (f-a)$ yield? I tried to search through convolution properties but I couldn't find an answer. Thank you. 212.98.136.42 (talk) —Preceding undated comment was added at 12:15, 13 August 2008 (UTC)

If you define, say, g(f) = G(kf), then

$g(f)*\delta (f-a)=g(f-a)=G(kf-ka).\,$

Hope that helps. siℓℓy rabbit (talk) 12:20, 13 August 2008 (UTC)

Yes I imagined it was $G(k(f-a))\,$

Thank you. 212.98.136.42 (talk) 12:34, 13 August 2008 (UTC)

Note that your $G(f)*\delta (f-a)=G(f-a)$ is something of an abuse of notation. Convolution is normally defined as an operator acting on functions, such that it would, for example, be meaningful to write $(G*\delta )(f)=G(f)$ or just $G*\delta =G$. (Indeed, this is basically the definition of the delta function.) Your expression only makes sense if one interprets $G(f-a)$ as a shorthand for the composite function $G\circ \tau _{a}$, where $\tau _{a}(f)=f-a$. Without such shorthand, it could be written as $G*(\delta \circ \tau _{a})=G\circ \tau _{a}$, with $\tau _{a}$ as defined above. It then clearly follows that $(G\circ q)*(\delta \circ \tau _{a})=(G\circ q)\circ \tau _{a}=G\circ q\circ \tau _{a}$, for all functions $q$, including the particular case $q(f)=kf$. —Ilmari Karonen (talk) 19:30, 13 August 2008 (UTC)

Marge Inoverra election

If, in a two-candidate race, a poll says candidate A has 45% of the vote, and candidate B has 55% of the vote, with a 5% margin of error, does that 5% apply to each candidate, so that A has 40-50% of the vote and B has 50-60% of the vote, or should the 5% be split by the two candidates to 2.5% each, for a range of 42.5-47.5% for candidate A and 52.5-57.5% for candidate B ? Also, how does it work with 3 or more candidates ? StuRat (talk) 15:27, 13 August 2008 (UTC)

The first case. The usual meaning of this would be that A has a 95% chance of getting 40-50% of the vote. See Margin of error for more details. Dmcq (talk) 22:26, 13 August 2008 (UTC)

Oh come on someone should have got at me for that :) What it really means is they asked a question on the vote to about a thousand people. Since about half said either way the standard deviation is sqrt(.45 * .55 /1000) or about 1/60. 95% is about two standard deviations or about 1/30 which is a bit over 3% but they just said 5% as it looks good and there's whole lots of problems with the procedure. To get a decent estimate they'd have to do stratified sampling and make sure their question was phrased just right. Even so there'd be lots of people who, for example, see the nice researcher as left or right wing and give the answer they think will make the researcher pleased. An opinion poll gives an answer to a question at the time it's run. As to how the actual vote will go, that may well just depend on something like the weather on the day. The pollsters aren't going to specify any tighter interval because they don't like people making fun of them after misunderstanding what they're doing. They do also sometimes make predictions and I suppose these will get better and better as they start using all sorts of other measures as well, e.g. pro and anti posts on the web, but I don't think you'll find margins of error on those yet. Dmcq (talk) 07:54, 14 August 2008 (UTC)