Why are simultaneous confidence intervals wider than the normal ones? Why are the simultaneous confidence intervals wider than the normal ones?
Aren't they made for securing that intervals equal amongst all the parameters, so separate intervals will be made for each parameter? Why does that make them wider?
 A: Consider two parameters and assume for simplicity that they are estimated independently. Say the sample mean and variance of a normal population which are estimates for the respective population quantities and are known to be indepedent. 
From basic statistics we know how to construct confidence intervals for $\mu$ and $\sigma^2$ based on our estimates. Since this is a normal distribution, we have two beautiful pivotal quantities, namely
$$\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1)\quad \text{and} \quad \frac{\left(n-1 \right) S^2}{\sigma^2} \sim \chi^2_{n-1} $$
Based on these distributions we can now find quantiles such that
$$ P \left[ -z_{0.975} < \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} < z_{0.975} \right] =0.95 \quad \text{and} \quad P\left[ \chi^2_{0.025}<\frac{\left(n-1 \right) S^2}{\sigma^2} < \chi^2 _{0.975} \right] = 0.95$$
Easy enough but there is a logical fallacy in these intervals, which you might have picked up. In the first interval I have assumed that $\sigma$ is known and this is clearly not the case, otherwise we wouldn't need the second interval! 
What we can do to resolve this contradiction is use the four inequalities to determine a confidence region for both $\mu$ and $\sigma^2$. Thus we will now be concerned with the probability
\begin{align} \label{eq:1} P\left[-z_{0.975} < \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} < z_{0.975} \ , \  \chi^2_{0.025} < \frac{\left(n-1 \right) S^2}{\sigma^2} < \chi^2 _{0.975}\right] \tag{1} \end{align}
With a slight reformulation of the above inequalities it is possible to determine the boundaries of $\mu$ and $\sigma^2$. Here is how the resulting region looks like

Pay no attention to the labels in the graph, these are just minor differences in the notation of the quantiles. You can now project onto the coordinate axes to obtain a confidence interval for both $\mu $ and $\sigma^2$. But we still do not know the level of this confidence region.
We have now arrived at the crux of the matter. My question to you is exactly that : what is the confidence level of this region? Surely we have used two 95% intervals so at first one might think that we are still at the 95% level. A moment of reflection will reveal however that this is incorrect. 
In this simplified setting the confidence level can be computed exactly using the independence $\bar{X}$ and $S^2$. Take a look at equation (\ref{eq:1}) and recall that for two independent events A and B, $P\left[ A \cap B \right] = P[A] P[B]$. Thus
$$  P\left[z_{0.975} < \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} < z_{0.975} \ , \  \chi^2_{0.025} < \frac{\left(n-1 \right) S^2}{\sigma^2} < \chi^2 _{0.975}\right] = 0.95^2 = 0.9025$$
And we have found out the hard way that two 95 % intervals do not mean that the joint confidence level, i.e. the confidence level of equation (\ref{eq:1}), is 0.95. By Bonferroni's inequality we can establish a lower bound - which is nearly attained in this case - but other than that not much can be said for the general case of non-independence.
In view of this, it is often preferable to take a conservative approach and specify simultaneous confidence intervals that have at least 0.95 coverage. Indeed, the concept of exact confidence interval no longer exists in simultaneous interval estimation. Tukey's, Scheffe's and even Bonferroni's procedures are all conservative in nature. For instance, in ANOVA when you want to test contrasts you will be able to build intervals that are at least of 0.95 level but you can never be sure what their exact level is. 
Of course, the intervals then are much wider than in the individual cases. It appears that we cannot do better, however, so better be safe than sorry.
Hope this helps.
A: Consider simultaneously estimating two parameters. We want to guarantee that, under repeated sampling, for some confidence level c and significance level $\alpha$, $(c = 1-\alpha )$% of our confidences interval pairs (one on each parameter) would both include the parameter.
Without loss of generality, let $\alpha$ = 0.05, c = 0.95. Let's construct single individual confidence intervals on each of them. The probability of each of these intervals containing their parameter is 0.95 under repeated sampling. Assuming independence, the probability of them both including their parameters under repeated sampling is 0.95*0.95 = 0.9025 < 0.95. If we wanted them both to include the true parameter values under repeated sampling 0.95 of the time, we would need $\alpha = \sqrt{0.95} = 0.975...$, which will obviously be a bigger interval on each parameter.
We will consider two levels of confidence here (we will assume that all the parameters we are measuring are the same for simplicity, so pretend we are simultaneous estimating the mean of two normal distributions in this discussion): one overall confidence level ($c_1$), and one confidence level for each parameter ($c_2$). $c_1$ is the long term probability of *every single one * of our confidence intervals successfully covering their respective parameters under repeated sampling, and is what we are directly interested in. If I were to say: "I want a 95% CI on these two parameters", the 95% = 0.95 corresponds to $c_1$. Once we have $c_1$, it falls to us to calculate $c_2$, which in our example here with two independent parameters is about 0.975. $c_2$ represents the "confidence level" we will have on each parameter. It has to be greater than if we were just estimating that parameter alone, and so will lead to a greater interval. 
As the number of parameters to simultaneously estimate increases, the confidence of getting each individual one right will also have to increase for static overall confidence, because as we have more parameters to estimate, we have more opportunities for failure under repeated sampling.
For example, consider estimating a million parameters. If we give them all 0.95 individual confidence intervals, there's no way we would be say that the set of all confidence intervals here will include the true parameters under repeated sampling 95% of the time under repeated sampling.
