Why is the exponential family so important in statistics? Why is the exponential family so important in statistics?
I was recently reading about the exponential family within statistics. As far as I understand, the exponential family refers to any probability distribution function that can be written in the following format (notice the "exponent" in this equation):

This includes common probability distribution functions such as the normal distribution, the gamma distribution, the Poisson distribution, etc. Probability distributions from the exponential family are often used as the "link function" in regression problems (e.g., in count data settings, the response variable can be related to the covariates through a Poisson distribution) - probability distribution functions that belong to the exponential family are often used due to their "desirable mathematical properties". For example, these properties are the following:

Why are these properties so important?
A) The first property is about "sufficient statistics". A "sufficient statistic" is a statistic that provides more information for any given data set/model parameter compared to any other statistic.
I am having trouble understanding why this is important. In the case of logistic regression, the logit link function is used (part of the exponential family) to link the response variable with the observed covariates. What exactly are the "statistics" in this case (e.g.. in a logistic regression model, do these "statistics" refer to the "mean" and "variance" of the beta-coefficients of the regression model)? What are the "fixed values" in this case?
B) Exponential families have conjugate priors.
In the Bayesian setting, a prior p(thetha | x) is called a conjugate prior if it is in the same family as the posterior distribution p(x | thetha). If a prior is a conjugate prior - this means that a closed form solution exists and numerical integration techniques (e.g., MCMC) are not required to sample the posterior distribution. Is this correct?
C) Is the third property essentially similar to the second property?
D) I don't understand the fourth property at all. Variational Bayes are an alternative to MCMC sampling techniques that approximate the posterior distribution with a simpler distribution - this can save computational time for high dimensional posterior distributions with big data. Does the fourth property mean that variational Bayes with conjugate priors in the exponential family have closed form solutions? So any Bayesian model that uses the exponential family does not require MCMC - is this correct?
References:

*

*Exponential family

*Sufficient statistic

*Conjugate prior

Great questions.  There are lots of ways to answer these. John Madden does an excellent job, but I'm going to crib a little bit from Ben's answer here regarding sufficient statistics.
The loss function for a Gaussian linear model (as Ben notes) is
$$ \ell_{\mathbf{y}, \mathbf{x}}(\boldsymbol{\beta}, \sigma)=-n \ln \sigma-\frac{1}{2 \sigma^{2}}\|\mathbf{y}-\mathbf{x} \boldsymbol{\beta}\|^{2}$$
In a lot of pedagogical material on fitting models, we would use the data to compute gradients of the loss and perform some sort of optimization routine.  Some code to do this might look like
def compute_loss(beta, X, y):
    
    number_rows = len(X)
    loss = 0
    
    # Potentially expensive!
    for i in range(number_rows):
        loss += (y[i] - X[i]@beta)**2
        
    loss/=number_rows
    
    return loss

If we have lots of data, then this loop (and any additional iterations over the data to compute gradients, for example) might be expensive to compute.  But, because we can write down sufficient statistics for the exponential family, we can improve our computation drastically.
As Ben writes, the loss can be rewritten as
$$
\begin{aligned}
&=-n \ln \sigma-\frac{1}{2 \sigma^{2}} \mathbf{y}^{\mathrm{T}} \mathbf{y}-\frac{1}{2 \sigma^{2}}\left(2 \boldsymbol{\beta}^{\mathrm{T}} \mathbf{T}_{1}-\boldsymbol{\beta}^{\mathrm{T}} \mathbf{T}_{2} \boldsymbol{\beta}\right) \\
&=h(\mathbf{y}, \sigma)+g_{\beta}\left(\mathbf{T}_{1}, \mathbf{T}_{2}, \sigma\right)
\end{aligned}
$$
Where $\mathbf{T}_{1} \equiv \mathbf{T}_{1}(\mathbf{x}, \mathbf{y}) \equiv \mathbf{x}^{\mathrm{T}} \mathbf{y}$ and $\mathbf{T}_{2} \equiv \mathbf{T}_{2}(\mathbf{x}, \mathbf{y}) \equiv \mathbf{x}^{\mathrm{T}} \mathbf{x}$.  We can thus compute these quantities one time and fit our models, as opposed to recomputing them at each update step.
Let's compare on a simulated problem.  I will assume $\sigma=1$ and does not require estimation for simplicity. I'll vary the number of observations and the number of covariates in the model and compare the time to optimize the loss assuming when using sufficient stats and when using a naive approach.  Shown below is a plot of the expected time to completion plus/minus a standard deviation.  We can see that as the data become larger in size, using sufficient statistics is advantageous.

Now, this whole answer is a bit of a straw man.  I've not shown you the results when not using sufficient statistics and not using loops (so maybe leveraging some linear algebra).  But the point about sufficient statistics remains.  That we can represent all the information in a sample with a single number, computed once, is a very valuable property to have.
Code to reproduce experiments:
import numpy as np
import pandas as pd
from scipy.optimize import minimize
from itertools import product
import matplotlib.pyplot as plt
import seaborn as sns
sns.set_theme(style="darkgrid")

# First, simulate big data!

def make_data(N, p):
    X = np.random.normal(size = (N,p))
    beta = np.random.normal(2, 2, size = p)
    y = X@beta + np.random.normal(size = N)
    return X,y


# Next, set up a loss function to optimize using loops
def solve_naive(N, p):
    X, y = make_data(N, p)
    def loss_and_grad(w):
        
        number_rows, number_columns = X.shape
        grads = np.zeros_like(w)
        loss=0
        for i in range(number_rows):
            res= (y[i] - X[i]@w)
            loss+= res**2/number_rows
            
            grads+= -2*res*X[i]/number_rows
        
        return loss, grads
    
    r=minimize(loss_and_grad, x0=np.zeros(p), jac=True)
    return loss_and_grad


# Next, set up a loss function to opimize which only uses sufficient statistics
def solve_sufficient(N, p):
    X, y = make_data(N, p)
        
    T1 = X.T@y
    T2 = X.T@X
    const = y.T@y
    n = len(X)
    
    def loss_and_grad(w):
        
        loss = const - (2*w@T1 - w@T2@w)
        
        grads = -2*T1 + (T2@w + T2.T@w)
        
        return loss/n, grads/n

    r=minimize(loss_and_grad, x0=np.zeros(p), jac=True)
    return loss_and_grad
    

# A helper function to time the optimization for various datasets
def time_to_optimize(N, p):
    


    naive_optimization_times = %timeit -o -n 10 solve_naive(N, p)
    suff_optimization_times = %timeit -o -n 10 solve_sufficient(N, p)
    
    suff = pd.DataFrame({'times':suff_optimization_times.timings,
                         'type': 'Sufficient Statistics'})
    
    naive = pd.DataFrame({'times':naive_optimization_times.timings,
                         'type': 'Naive'})
    
    df = pd.concat((suff, naive))

    
    df['N'] = N
    df['p'] = p
    
    return df

if __name__ == '__main__':

    Ns = [1_000, 10_000, 100_000]
    ps = [10, 100, 250]

    prods = product(Ns, ps)
    frames = []

    for N, p in prods:

        frames.append(time_to_optimize(N, p))
        

    df = pd.concat(frames).reset_index(drop=True)

    fig, ax = plt.subplots(dpi = 240, figsize = (8, 5))
    grid = sns.lineplot(data=df, x='N', y='times', hue='type', style='p')

    grid.set(xscale="log", yscale="log", xlabel = 'Number of Observations', ylabel='Execution Time (Seconds)')
    grid.legend(loc='best', prop = {'size':6})





