[R] Fwd: Find maxima of a function
Bert Gunter
bgunter.4567 at gmail.com
Sun Aug 27 19:43:15 CEST 2017
Just a quick note:
"I have tried to use optimr but it gives me the local maxima, now I am
struck with the problem of how to get the global maxima"
This is a profound misunderstanding of how numerical optimization
works. **All** numerical optimizers can provide only "local" maxima
(the exact meaning of "local" may vary). It is mathematically
impossible to numerically obtain "global" maxima and know that you
have done so.*
-- Bert
*Unless, of course, you can prove analytically that only global maxima
exist or some other such supplementary mathematical information.
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Aug 26, 2017 at 6:39 AM, niharika singhal
<niharikasinghal1990 at gmail.com> wrote:
> Hi,
>
> Thanks for your mail, and time
>
> It is not working for some arguments, when mean value is like >6.
>
>
> case
>
> mc0 <- c(0.08844446,0.1744455,0.1379778,0.1209769,0.1573065,0.
> 1134463,0.2074027)
>
> rv <-UnivarMixingDistribution(Norm(486.4255, 53.24133),
>
> Norm(664.0713, 3.674773),
>
> Norm(669.0484, 4.101381),
>
> Norm(677.1753, 4.869985),
>
> Norm(683.2635, 7.288175),
>
> Norm(727.6229, 37.64198),
>
> Norm(819.2011, 57.06655),
>
> mixCoeff=mc0/sum(mc0))
>
> plot(rv, to.draw.arg="d")
>
>
> I am getting 731.1345 from the code you have provide
>
>
> It is part of a code, so it was difficult to write a reproducible code
>
> I have tried to use optimr but it gives me the local maxima, now I am
> struck with the problem of how to get the global maxima
>
>
> On Sat, Aug 26, 2017 at 3:12 PM, Ismail SEZEN [via R] <
> ml+s789695n4744993h60 at n4.nabble.com> wrote:
>
>>
>> > On 26 Aug 2017, at 14:18, Ulrik Stervbo <[hidden email]
>> <http:///user/SendEmail.jtp?type=node&node=4744993&i=0>> wrote:
>> >
>> > Please keep the list in cc.
>> >
>> > Sorry, it didn't work as expected. Maybe someone else have an
>> appropriate
>> > solution.
>> >
>> > Best,
>> > Ulrik
>> >
>> > On Sa., 26. Aug. 2017, 12:57 niharika singhal <[hidden email]
>> <http:///user/SendEmail.jtp?type=node&node=4744993&i=1>>
>> > wrote:
>> >
>> >> Hi
>> >>
>> >> Thanks for you mail,
>> >> I really appreciate your time on my problem
>> >>
>> >> I have posted this problem on
>> >>
>> >>
>> >> https://stats.stackexchange.com/questions/299590/to-find-max
>> ima-for-gaussian-mixture-model
>> >>
>> >>
>> >> The plot I am getting using UnivarMixingDistribution from distr package
>> in
>> >> R
>> >>
>> >> code is
>> >>
>> >> mc0= c(0.1241933, 0.6329082 <06329%20082>, 0.2428986 <02428%20986>)
>> >> rv
>> >> <-UnivarMixingDistribution(Norm(506.8644,61.02859),Norm(672.8448,9.149168),Norm(
>>
>> >> 829.902,74.84682), mixCoeff=mc0/sum(mc0))
>> >> plot(rv, to.draw.arg="d")
>> >>
>> >> I want output around 672 in first case and in 2nd case around 2.1
>> >> according to the plot.
>> >> your code will not work in both the scenario
>> >>
>> >> Regards
>> >> Niharika Singhal
>> >>
>> >>
>> >> On Sat, Aug 26, 2017 at 12:47 PM, Ulrik Stervbo <[hidden email]
>> <http:///user/SendEmail.jtp?type=node&node=4744993&i=2>>
>> >> wrote:
>> >>
>> >>> Hi,
>> >>>
>> >>> I once found this somewhere on stackoverflow:
>> >>>
>> >>> values <- rnorm(20, mean = c(2.15,2.0,2.9), sd = c(0.1,0.1,0.1))
>> >>>
>> >>> v_dens <- density(values)
>> >>> v_dens_y <- v_dens$y
>> >>>
>> >>> r <- rle(v_dens_y)
>> >>> # These functions ignore the extremes if they are the first or last
>> point
>> >>> maxima_index <- which(rep(x = diff(sign(diff(c(-Inf, r$values,
>> -Inf))))
>> >>> == -2, times = r$lengths))
>> >>> minima_index <- which(rep(x = diff(sign(diff(c(-Inf, r$values,
>> -Inf))))
>> >>> == 2, times = r$lengths))
>> >>>
>> >>> plot(v_dens_y)
>> >>>
>> >>> HTH
>> >>> Ulrik
>> >>>
>> >>>
>> >>> On Sat, 26 Aug 2017 at 11:49 niharika singhal <
>> >>> [hidden email] <http:///user/SendEmail.jtp?type=node&node=4744993&i=3>>
>> wrote:
>> >>>
>> >>>> I have a Gaussian mixture model with some parameters
>> >>>>
>> >>>> mean=(506.8644,672.8448,829.902)
>> >>>>
>> >>>> sigma=(61.02859,9.149168,74.84682)
>> >>>>
>> >>>> c=(0.1241933, 0.6329082 <06329%20082>, 0.2428986 <02428%20986>)
>> >>>>
>> >>>> And the plot look something like below.[image: enter image
>> description
>> >>>> here]
>> >>>> <https://i.stack.imgur.com/4uUQ9.png>
>> >>>>
>> >>>> Also, if I change my parameters to
>> >>>>
>> >>>> mean=(2.15,2.0,2.9)
>> >>>>
>> >>>> sigma=(0.1,0.1,0.1)
>> >>>>
>> >>>> c=(1/3,1/3,1/3)
>> >>>>
>> >>>> Then plot would change to[image: enter image description here]
>> >>>> <https://i.stack.imgur.com/kESYX.png>
>> >>>>
>> >>>> Is there any way to find the maxima. I have tried Newton's method but
>> it
>> >>>> gave me the wrong output.
>> >>>>
>> >>>> Like in general some common solution, which would work on all the
>> cases,
>> >>>> is
>> >>>> needed.Can someone suggest me how can I achieve this
>> >>>>
>> >>>> Thanks in advance
>> >>>>
>> >>>> Niharika Singhal
>> >>>>
>> >>>> [[alternative HTML version deleted]]
>> >>>>
>> >>>> ______________________________________________
>> >>>> [hidden email]
>> <http:///user/SendEmail.jtp?type=node&node=4744993&i=4> mailing list --
>> To UNSUBSCRIBE and more, see
>> >>>> https://stat.ethz.ch/mailman/listinfo/r-help
>> >>>> PLEASE do read the posting guide
>> >>>> http://www.R-project.org/posting-guide.html
>> >>>> and provide commented, minimal, self-contained, reproducible code.
>> >>>>
>> >>>
>> >>
>>
>> See the solution below if it helps. Also please, create a minimal
>> reproduciple example as I did. So, we can investigate easily on your real
>> problem.
>>
>> set.seed(1)
>> m <- c(506.8644, 672.8448, 829.902)
>> sig <- c(61.02859, 9.149168, 74.84682)
>> x1 <- density(rnorm(500, mean = m, sd = sig))
>> plot(x1)
>>
>> fun <- splinefun(x = x1$x, y = x1$y, method = "n")
>> optimize(fun, interval = range(x1$x), maximum = TRUE)
>>
>>
>> m <- c(2.15, 2.0, 2.9)
>> sig <- c(0.1,0.1,0.1)
>> x2 <- density(rnorm(500, mean = m, sd = sig))
>> plot(x2)
>>
>> fun <- splinefun(x = x2$x, y = x2$y, method = "n")
>> optimize(fun, interval = range(x2$x), maximum = TRUE)
>>
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>>
>>
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> [[alternative HTML version deleted]]
>
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