How is MAP 'not invariant to reparametrization'? I was watching a lecture on coursera on 'Bayesian Methods on Machine Learning' and I came across a statement that: MAP(Maximum a posteriori) is not invariant to reparametrization. I didn't quite understand:


*

*What does reparametrization mean here and why is reparametrization important?

*How is MAP 'not invariant to reparametrization'? 

*Why 'not being invariant to reparametrization' a problem and How do Conjugate priors help solve this problem?
Please explain the answers in an easy and intuitive way. I looked for other similar questions to mine and they are way too mathematical and I don not have a solid theoretical foundation in bayesian statistics.
PS: Please do not mark this question as a duplicate. I have read other similar questions on StackOverflow and other sites, however, those questions don't answer my questions completely and clearly.
Edit: Even after writing a special note that I have viewed every possible question similar to this, and those do not answer my questions, This question was marked duplicate.
 A: MAP estimation relies at its heart using optimization of the posterior (ignoring the constant term) across the values of your parameters $\theta$.  Mathematically: $\theta^{MAP} = argmax_{\theta} p(x|\theta)p(\theta)$.  Now suppose we would like to reparameterize $k= log(1-exp(\theta))$ and solve for k.  Well quite simply the maximization problem will not give the same results even solving back for k i.e. $k^{MAP}= argmax_k p(x|k)p(k)$ and then $ \theta^{new} = log(1-e^{k^{MAP}})\neq \theta^{MAP}$ for most combinations of likelihood and priors in the original model.
Why reparameterization like the above (meaning do algebraic manipulation to get a nicer form for the parameter of interest)?  Well sometimes using the above reparameterization have nice computational properties because for normals for example exponential components can get very large very quickly and the computer can overflow. Checking probabilities in the distribution can be very important to check how strong your model is.
Lastly being not invariant to reparameterization makes a method less attractive. If you can easily manipulate your problem into a simpler or more computer-friendly method there can be large gains to efficiency. This efficiency might mean that things can be estimated more accurately or much more quickly. As to the last part of your question 3, I'm not completely sure.  Perhaps the posterior being in the same family as the prior ensures that when you "retranslate" back to your original parameter you will get the same answer had you not reparameterized, but I'm not completely sure and I'd have to see it worked out to know for certain.
