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Exercise 2.5 Hi, i am fairly new on these forums, so please let me know, if information or formulation is lacking.

My question is related to the picture i have linked. My problem is i really don't understand how the likelihoods of $\theta$ is derived and my intuition of likelihood functions is lacking a lot. Could anyone explain to me how i would go on about solving these problems? I find (c), (d), (e) especially hard to grasp.

Thanks a lot :)

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  • $\begingroup$ hey @Chris, check this out and maybe consider adding the 'self-study' tag stats.stackexchange.com/tags/self-study/info $\endgroup$
    – Taylor
    Sep 8, 2017 at 16:54
  • $\begingroup$ Do you any definition of likelihood? Can you give your solutions for (a) and (b)? $\endgroup$
    – Juho Kokkala
    Sep 8, 2017 at 18:02
  • $\begingroup$ I know the definition of likelihood, my solution for (a) is the product of all the normal density functions of each attribute with a mean of $\theta$ and the SD of the given sample. (b) is then the same solution but with the standard deviation given as $\frac{SD(x)}{\sqrt{n}}$, since we only got the sample mean. I am getting a bit lost at (c), (d) and (e). $\endgroup$
    – Chris
    Sep 8, 2017 at 18:19
  • $\begingroup$ NB The q. does say "$\sigma^2$ is known at the observed sample variance". Please edit your question to show your working & where you're stuck. $\endgroup$
    – Scortchi - Reinstate Monica ♦
    Sep 9, 2017 at 19:29

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likelihood is actually the joint density function but the interpretation is a little different. Consider a sample from any distribution whose parameter(s) are unknown. For simplicity let's say Normal(a,1) i.e, mean =a ; variance =1. Now the maximum likelihood estimate of a means that for what value of a are you most likely to draw the sample that you have. Meaning that after you estimate a, the distribution is identified so you can always draw a random sample. The probability that you draw the sample you have is maximised with the mle. Now it's obvious that you will actually never get exactly the sample you have, since it's a continuous distribution so the probability of the random variable taking a particular value is always 0. Here you are maximising the Probability of drawing the sample.

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