Flaws in Frequentist Inference I have problem to understanding the following example.

(1) After the next day that the glitch discovered what can tell about the observation? $X_i\nsim N(\mu,1)$ or just $X_i\sim N(\mu_2,1)$. Some observation are from $N(\mu,1)$ and others not. What can we tell about all observation. Am I wrong somthing? Why not using Truncated normal? 
(2) What did exactly bayesian inference do in this situation? I think ,by considering $\mu$ to be random variable, it control (model) the glitch, but it considering $X\sim N(\mu , 1)$
that I am doubting to accept this. 
(3) At All, is the conversation "3.3) Flaws in frequentist inference" valid? Is this fair? and
with out exaggeration ?
source:Computer Age Statistical Inference, by Bradley Efron and Trevor Hastie, Page 31.
 A: I think this is exaggerated language. Both frequentist and Bayesian have their merits, and statisticians routinely rely on both types in their work. To answer your questions: 


*

*We can still consider $X \sim N(\mu, 1)$. However, we are not observing $X$, but $X' = \min(100, X)$, which is another Random Variable. 


2,3. Now, the authors are saying $\mathbb{E}(X') \neq \mu$, even though $\mathbb{E}(X) = \mu$. Is this a limitation in frequentist inference? Perhaps. Many statistical inference techniques (e.g. confidence intervals, p-value, hypothesis testing) require that our estimator ($\hat{\mu}$) to be a consistent (roughly speaking: asymptotically unbiased) estimate of $\mu$. Now, the thing about Bayesian inference is that it does not care (much) about biasedness. It doesn't solve the frequentist problem. It just provides another perspective on it. 
The particular example given, however, is specially constructed to be paradoxical. Since we know that no values of $x_i > 100$ are observed, the mean observed $\bar{x}'=\sum_i x'_i/n$, is an unbiased estimate of $\mu$. But this is only due to an idiosyncracy of this setup, since in this particular case, $\bar{x}'=\sum_i x'_i/n = \sum_i x_i/n = \bar{x}$. However, in general, $\mathbb{E}(\bar{x}') \neq \mu$, even though $\mathbb{E}(\bar{x}) = \mu$. 
A Bayesian, however, can be more flexible. First, she can define $S$ to be the event that "a randomly drawn sample contains no $x'_i=100$". Then she can say, $\mathbb{E}(\bar{x}' | S) = \mathbb{E}(\bar{x}|S) = \mu$. Note that the precise statement is quite delicate. If the sample did in fact contain one $x'_i = 100$, but we are excluding that observation, or if we are repeating the sampling procedure, and only selecting one which doesn't contain any $x'_i=100$, this is not the event specified by $S$. 
This is, in some way, the limitation of frequentism. Conditioning on an event like $S$ simply does not fit in with frequentist philosophy, since it is concerned with the long term behaviour of a procedure or estimator. 
On the other hand, the fact that in this particular case, the Bayesian has an unbiased estimate is something of a fluke. Bayesianism does not in the end solve the questions frequentist inference tries to address. 
#### Edit ####
I just realized I actually got tricked by the paradox. Using an event like $S$ does not solve the problem, as it still implies $x_i < 100$ for all observed $i$. Ultimately, I think the answer is that the Bayesian simply is not interested in $\mathbb{E}(X')$, and the frequentist cannot consider $\mathbb{E}(\mu|X'=x')$. Thus the fact that the estimate is biased or not is irrelevant to the Bayesian. A Bayesian may of course still be interested in analysing the properties of an estimator, but that will then make him a frequentist. 
A: It is a bit sad to see printed such carelessly written prose.
Consider the phrase 

"For any prior density $g(\mu)$, the posterior density $g(\mu\mid x)=
g(\mu)f_{\mu}(x)/f(x)$ ....depends only on the data actually
  observed..."

-while the mathematical formula in the above same sentence shows that the posterior density depends on the data and on the prior density (and let's not discuss how do we determine $f_{\mu}(x)$ and $f(x)$).
Second, we have here a posterior out of sample information: we have discovered that the sample is censored. Bayesians have championed the formal and transparent inclusion of out-of-sample information in our estimation procedures, so this example should have been used to show how we could incorporate the discovered sample censoring into the estimation.
On the contrary, the passage almost advises us to ignore this information, since it chooses to end by using the example of a flat prior. that, we learn,  would yield posterior expectation $92$... But it would be a serious mistake to use it because, Bayesian, frequentist or whatever, it is always a serious mistake to ignore the facts. But the passage ends almost marveling at and celebrating, that we would get $92$, 

irrespective of whether or not the glitch would have affected readings
  above 100.

The correct answer is $42$, by the way.
A: I am a Bayesian, but I find these kinds of criticisms against "frequentists" to be overstated and unfair.  Both Bayesians and classical statisticians accept all the same mathematical results to be true, so there is really no dispute here about the properties of the various estimators.  Even if you are a Bayesian, it is clearly true that the sample mean is no longer an unbiased estimator (the very concept of "bias" being one that conditions on the unknown parameter).  So first of all, the frequentist is correct that the sample mean is not an unbiased estimator (and any sensible Bayesian would have to agree with this given the assumed distributions).  Secondly, if a frequentist actually encountered this situation, they would almost certainly update their estimator to reflect the censoring mechanism in the data.
It is entirely possible for the frequentist to use an estimator that is unbiased, and which reduces down to the sample mean in the special case where there is no censored data.  Indeed, most standard frequentist estimators would have this property.  So, although the sample mean is indeed a biased estimator in this case, the frequentist could use an alternative estimator that is unbiased, and which happens to give the same estimate as the sample mean for this particular data.  Therefore, as a practical matter, the frequentist can happily accept the estimate from the sample mean is the correct estimate from this data.  In other words, there is absolutely no reason that the Bayesian needs to "come to the rescue" --- the frequentist will be able to accomodate the changed information perfectly adequately.

More detail: Suppose you have $m$ non-censored data points $x_1,...,x_m$ and $n-m$ censored data points, which are known to be somewhere above the cut-off $\mu_* = 100$.  Given the underlying normal distribution for the pre-censored data values, the log-likelihood function for the data is:
$$\ell_\mathbb{x}(\mu) = \sum_{i=1}^m \ln \phi (x_i-\mu) + (n-m) \ln (1 - \Phi(\mu_*-\mu)).$$
Since $\ln \phi (x_i-\mu) = - \tfrac{1}{2}(x_i-\mu)^2+\text{const}$, differentiating gives the score function:
$$\frac{d \ell_\mathbb{x}}{d \mu}(\mu) = m (\bar{x}_m - \mu)
+ (n-m) \cdot \frac{\phi(\mu_*-\mu)}{1 - \Phi(\mu_*-\mu)}.$$
so the MLE is the value $\hat{\mu}$ that solves:
$$\bar{x}_m = \hat{\mu} + \frac{n-m}{m} \cdot \frac{\phi(\mu_*-\hat{\mu})}{1 - \Phi(\mu_*-\hat{\mu})}.$$
The MLE will generally be a biased estimator, but it should have other reasonable frequentist properties, and so it would probably be considered a reasonable estimator in this case.  (Even if the frequentist is looking for an improvement, like a "bias corrected" scaled version of the MLE, it is likely to be an other estimator that is asymptotically equivalent to the MLE.)  In the case where there is no censored data we have $m=n$, so the MLE reduces to $\hat{\mu} = \bar{x}_m$.  So in this case, if the frequentist used the MLE, they will come to the same estimate for non-censored data as if they were using the sample mean.  (Note here that there is a difference between an estimator (which is a function) and an estimate (which is just one or a few output values from that function).
A: It's worth noting that there is nothing that prevents Frequentist analysis from saying 
"Conditional on none of your data being censored, $\hat \mu$ is equal to $\bar x$ and will be unbiased. Conditional on some of your data being censored, the MLE estimator $\hat \mu$ is no longer equal to $\bar x$ and has some bias". 
Of course, marginalizing over whether there is censored data means that this whole framework has bias, but there's nothing that prevents Frequentists from making conditional statements. 
