The formula for the mean of a pdf is in the tables. I can calculate the mode, but am unsure how to work out the median. does anyone know how to do this? This is in relation to questions asking you to find the bayesian estimate under squared error loss, absolute error loss etc.
being pedantic jayney states that s/he can find the mode of a pdf , well there aint no such thing pdf s are continuous so no mode only on discrete distributions like the poisson is there a mode i.e. only on what are called p.m.f.s probability mass functions .Hope this doesnt annoy everyone too much , like I said being pedantic.
Actually, that's not correct. The mode of a pdf is the value of its global maximum. Normal, gamma etc are cont. distributions and these have modal values
short cuts For calculating the median, don't forget that, if the cumulative distribution is defined, then we can short cut P(X < M) = 0.5 by solving: F(M) = 0.5 Since F is in the Tables for many of the distributions (eg Pareto, exponential, uniform) then this usually the quickest approach. Note for the standard normal, or chi-square distribution, you can use the tabulated values in the Tables to find this median value. Note that for the non-standard normal or the lognormal, you need to standardise first and then use the tabulated values. Eg 0.5 = P(X < M) = P(Z < [M - mu]/sigma) if X is N(mu, sigma^2) 0.5 = P(X < M) = P(Z < [ln M - mu]/sigma if X is logN(mu sigma^2) Note for the Gamma distribution, big F is undefined. You need to convert to a chi-square distribution. Eg 0.5 = P(X < M) = P(Y < 2 lambda M) where Y is chi-square with 2 alpha degrees of freedom and X is Gamma (alpha, lambda) Re the mode, this can be calculated for many continuous distributions. I've just looked it up in my dictionary of mathematics and it defines it as the maximum value in a probability distribution. So, take the pdf, differentiate and set to zero. Differentiate again to check for a max. Anna
An even more obvious short cut for the median in a normal distribution is Mean = Mode = Median Because of symmetry.
I stand corrected Suppose I was big time wrong there , but I still dont like the idea of a mode of a continuous dist just seems wrong , there is only interval probabilities after all .
I think you have to accept that the mode is ok for a continuous distribution as long as you define "mode" in a way that makes it acceptable. If you define it as "the most popular value", it's not going to work! However, if you define it as "the point of global maximum" then it's fine.
