What is the difference between converges in probability and distribution What is the difference between the statements:
converges in probability
and
converges in distribution
I am looking for a more intuitive explanation and simple formulas rather than abstract mathematical definitions. (I have no knowledge of measure theory or other abstract mathematical concepts).
I didn't find my answer in this topic: Convergence in probability and distribution 
 A: Trying my best to use only simple formulas and be intuitive. Consider a sequence of random variable $X_1, X_2, \ldots $, and another random variable $X$.
Convergence in distribution. For each $n$, let us take realizations (generate many samples) from $X_n$, and make a histogram. So we now have a sequence of histograms, one for each $X_n$. If the histograms in the sequence look more and more like the histogram of $X$ as $n$ progresses, we say that $X_n$ converges to $X$ in distribution (denoted by $X_n \stackrel{d}{\rightarrow} X$).
Convergence in probability. To explain convergence in probability, consider the following procedure. 
For each $n$, 


*

*repeat many times:


*

*Jointly sample $(x_n, x)$ from $(X_n, X)$.

*Find the absolute difference $d_n = |x_n - x|$.


*We have got many values of $d_n$. Make a histogram out of these. Call this histogram $H_n$.


If the histograms $H_1, H_2, \ldots $ become skinnier, and concentrate more and more around 0, as $n$ progresses, we say that $X_n$ converges to $X$ in probability (denoted by $X_n \stackrel{p}{\rightarrow} X$).
Discussion Convergence in distribution only cares about the histogram (or the distribution) of $X_n$ relative to that of $X$. As long as the histograms look more and more like the histogram of $X$, you can claim a convergence in distribution. By contrast, convergence in probability cares also about the realizations of $X_n$ and $X$ (hence the distance $d_n$ in the above procedure to check this). 

$(\dagger)$ Convergence in probability requires that the probability that the values drawn
  from $X_n$ and $X$ match (i.e., low $d_n = |x_n - x|$) gets higher and higher as
  $n$ progresses. 

This is a stronger condition compared to the convergence in distribution. Obviously, if the values drawn match, the histograms also match. This is why convergence in probability implies convergence in distribution. The converse is not necessarily true, as can be seen in Example 1.
Example 1. Let $Z \sim \mathcal{N}(0,1)$ (the standard normal random variable). Define $X_n := Z$ and $X := -Z$. It follows that $X_n \stackrel{d}{\rightarrow} X$ because we get the same histogram (i.e., the bell-shaped histogram of the standard normal) for each $X_n$ and $X$. Now, it turns out that $X_n$  does not converge in probability to $X$. This is because $(\dagger)$ does not hold. The reason is that realizations of $X_n$ and $X$ will never match, no matter how large $n$ is. Yet, they share the same histogram!
Example 2. Let $Y, Z \sim \mathcal{N}(0,1)$. Define $X_n := Z + Y/n$, and $X := Z$. It turns out that  $X_n \stackrel{p}{\rightarrow} X$ (and hence $X_n \stackrel{d}{\rightarrow} X$). Read the text at $(\dagger)$ again. In this case, the values drawn from $X_n$ and $X$ are almost the same, with $x_n$ corrupted by the added noise, drawn from $Y/n$. However, since the noise is weaker and weaker as $n$ progresses, eventually $Y/n=0$, and we get the convergence in probability.
Example 3. Let $Y, Z \sim \mathcal{N}(0,1)$. Define $X_n := Z + n Y$, and $X := Z$. We see that $X_n$ does not converge in distribution. Hence, it does not converge in probability.
A: Intuitively, convergence in probability means the random variables get close to a nonrandom constant, and convergence in distribution means that it gets close to another random variable. Closeness will mean different things in each situation. 
Example:
Say $X_1, \ldots, X_N$ are independent and identically distributed with mean $\mu$ (we don't have to assume which distribution they come from). By the Weak Law of Large Numbers
$$
\bar{X}_n \overset{p}{\to} \mu
$$
as $n \to \infty$. Here $\mu$ is a constant, and this result is useful because we know if we have a large dataset, we can effectively "recover" the true and unknown parameter $\mu$.
Also, by the Central Limit Theorem, if in addition to the above assumptions we assume the variance of each $X_i$ is $\sigma^2$, then 
$$
\sqrt{n}\bar{X}_n \overset{D}{\to} \text{Normal}(\mu, \sigma^2)
$$
as $n \to \infty$. Notice that the right hand side is a random variable. This is useful for making confidence intervals, and quantifying the uncertainty about estimates of $\mu$. If we have a large dataset, we can be justified in using those $z_{\alpha/2}$ standard Normal quantiles that are commonly found in CI formulas. Moreover, this works even if we don't make any assumptions about what probability distribution is generating the data.
Note that $\{\sqrt{n}\bar{X}_n\}$ is a different sequence than $\{\bar{X}_n\}$. We need a weaker type of convergence for the former because scaling it by bigger and bigger $\sqrt{n}$ increases the variance by an increasing $n$; however this "balances out" with the shrinking variance of $\bar{X}_n$. The net result is that the sequence of products coalesces around a random variable for which many formulas are known.
Even though convergence in probability for one sequence always implies convergence in distribution for the same sequence (this is mentioned in other answers), in this particular example, we have convergence in distribution for the scaled sequence implying convergence in probability for the sequence of un-scaled sample averages (kind of reversing the order of implication). I mention this because CLTs are often more valuable than results about convergence in probability. This is because if we know how fast we can scale up the sequence of estimators and still have it converge, this tells us something about the convergence "rate." In this particular example, the CLT tells us that $\bar{X}_n$ is "root-n consistent" (not to be confused with regular consistency, which is mentioned in other answers).
A: I'm not going to repeat the mathematical definitions here and here (or in the other answers). 
At a high level, I think it's helpful to think in terms of samples. The concepts of stochastic convergence are used to study how samples behave as the sample size increases. 
Individually,


*

*Convergence in probability is a weak statement to make. You may have seen this property being invoked when discussing the consistency of an estimator or by the Weak Law of Large Numbers. I think it's better to focus on how you can use this concept rather than the mathematical definition: let's say you're using your samples to estimate some true property in the wider population and you collect more and more samples. If your estimate converges in probability to the true value, then you have a consistent estimator (and that's a good thing!) because your estimate is now probably the true value.

*Convergence in distribution is an even weaker statement to make. It basically says that the Cumulative Distribution Functions of random variables converge (i.e. get more and more similar in their shapes). It doesn't discuss actual probabilities.
Relating the two concepts,


*

*Convergence in probability (stronger) implies convergence in distribution (weaker). Under certain conditions, the opposite relationship may hold as well.

