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[likelihood] solved standard deviation problem

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This commit is contained in:
Jan Benda 2021-01-18 18:58:37 +01:00
parent c5d5d94d1b
commit 51da7aa567
3 changed files with 12 additions and 7 deletions
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3
likelihood/exercises/likelihood-1.tex
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@ -37,10 +37,11 @@ of the standard deviation.
\end{parts} \end{parts}
\begin{solution} \begin{solution}
\lstinputlisting{mlestd.m} \lstinputlisting{mlestd.m}
\lstinputlisting[language={}]{mlestd.out}
\includegraphics[width=1\textwidth]{mlestd}\\ \includegraphics[width=1\textwidth]{mlestd}\\
The more data the smaller the product of the probabilities ($\approx The more data the smaller the product of the probabilities ($\approx
p^n$ with $0 \le p < 1$) and the smaller the sum of the logarithms p^n$ if $0 \le p < 1$) and the smaller the sum of the logarithms
of the probabilities ($\approx n\log p$, note that $\log p < 0$). of the probabilities ($\approx n\log p$, note that $\log p < 0$).
The product eventually gets smaller than the precision of the The product eventually gets smaller than the precision of the

16
likelihood/exercises/mlestd.m
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@ -6,16 +6,19 @@ for k = 1:length(ns)
fprintf('\nn=%d\n', n); fprintf('\nn=%d\n', n);
% draw random numbers: % draw random numbers:
x = randn(n,1)*sigma+mu; x = randn(n,1)*sigma+mu;
fprintf(' mean of the data is %.2f\n', mean(x)) datamean = mean(x);
fprintf(' standard deviation of the data is %.2f\n', std(x)) fprintf(' mean %.3f\n', datamean)
% we assume the mean to be given, therefore dof=n:
fprintf(' standard deviation %.3f\n', std(x, 1))
% standard deviation as parameter: % standard deviation as parameter:
psigs = 1.0:0.01:3.0; psigs = 1.0:0.001:3.0;
% matrix with the probabilities for each x and psigs: % matrix with the probabilities for each x and psigs:
lms = zeros(length(x), length(psigs)); lms = zeros(length(x), length(psigs));
for i=1:length(psigs) for i=1:length(psigs)
psig = psigs(i); psig = psigs(i);
p = exp(-0.5*((x-mu)/psig).^2.0)/sqrt(2.0*pi)/psig; % same mean as for the standard deviation of the data:
p = exp(-0.5*((x-datamean)/psig).^2.0)/sqrt(2.0*pi)/psig;
lms(:,i) = p; lms(:,i) = p;
end end
lm = prod(lms, 1); % likelihood lm = prod(lms, 1); % likelihood
@ -24,8 +27,9 @@ for k = 1:length(ns)
if length(maxlm) > 1 if length(maxlm) > 1
maxlm = maxlm(1); maxlm = maxlm(1);
end end
fprintf(' likelihood: standard deviation of the data is %.2f\n', maxlm) maxloglm = psigs(loglm==max(loglm));
fprintf(' log-likelihood: standard deviation of the data is %.2f\n', psigs(loglm==max(loglm))) fprintf(' likelihood: standard deviation %.3f\n', maxlm)
fprintf(' log-likelihood: standard deviation %.3f\n', maxloglm)
% plot likelihood of standard deviation: % plot likelihood of standard deviation:
subplot(2, 2, 2*k-1); subplot(2, 2, 2*k-1);

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likelihood/exercises/mlestd.pdf
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