[PYTHON] KL divergence between normal distributions

Introduction

I was not sure about the KL divergence that appears in the EM algorithm, so I would like to get an image by finding the KL divergence between the normal distributions.

KL divergence

Kullback-Leibler divergence (KL divergence, KL information amount) is a measure of how similar the two probability distributions are. The definition is as follows.

KL(p||q) = \int_{-\infty}^{\infty}p(x)\ln \frac{p(x)}{q(x)}dx

There are two important characteristics. The first is that it will be 0 for the same probability distribution.

KL(p||p) = \int_{-\infty}^{\infty}p(x)\ln \frac{p(x)}{p(x)}dx
         = \int_{-\infty}^{\infty}p(x)\ln(1)dx
         = 0

The second is that it will always be a positive value, including 0, and the more dissimilar the probability distributions, the larger the value. Let's look at these characteristics using an example of a normal distribution.

normal distribution

The probability density functions p (x) and q (x) of the normal distribution are defined as follows.

p(x) = N(\mu_1,\sigma_1^2) = \frac{1}{\sqrt{2\pi\sigma_1^2}} \exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right) \\
q(x) = N(\mu_2,\sigma_2^2) = \frac{1}{\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)

KL divergence between normal distributions

Find the KL divergence between the above two normal distributions. The calculation is omitted.

\begin{eqnarray}

KL(p||q)&=& \int_{-\infty}^{\infty}p(x)\ln \frac{p(x)}{q(x)}dx \\
        &=& \cdots \\
        &=& \ln\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{\sigma_1^2+(\mu_1-\mu_2)^2}{2\sigma_2^2} - \frac{1}{2}
\end{eqnarray}

Since it is difficult to understand if there are four variables, let $ p (x) $ be the standard normal distribution $ N (0,1) $ with mean 0 and variance 1.

p(x) =N(0,1)= \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right) 

When the mean is a variable

First, set the standard deviation of $ q (x) $ to $ \ sigma_2 $ and set only the mean $ \ mu_2 $ as variables.

q(x) =N(\mu_2,1)= \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{(x-\mu_2)^2}{2}\right) 

The KL divergence at this time is

\begin{eqnarray}

KL(p||q) &=& \ln\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{\sigma_1^2+(\mu_1-\mu_2)^2}{2\sigma_2^2} - \frac{1}{2} \\
         &=& \ln\left(\frac{1}{1}\right) + \frac{1^2+(\mu_1-0)^2}{2*1^2} - \frac{1}{2} \\
         &=& \frac{\mu_2^2}{2}
\end{eqnarray}

It will be. \mu_2The value of-Probability distribution when incremented by 1 from 4 to 4$q(x)And KL divergenceKL(p||q)$The value of is as follows.

KL_μ.gif

The orange line on the left is $ q (x) $ when the average $ \ mu_2 $ is changed. The figure on the right is the figure when the average $ \ mu_2 $ is taken on the x-axis. The blue line is the analytical solution, and the orange dot is the current KL divergence value. It was confirmed that the KL divergence becomes 0 when $ p (x) $ and $ q (x) $ match exactly, and increases as the distance increases.

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation

#normal distribution
def gaussian1d(x,μ,σ):
    y = 1 / ( np.sqrt(2*np.pi* σ**2 ) )  * np.exp( - ( x - μ )**2  / ( 2 * σ ** 2 ) )
    return y

#Normal distribution KL divergence
def gaussian1d_KLdivergence(μ1,σ1,μ2,σ2):
    A = np.log(σ2/σ1)
    B = ( σ1**2 + (μ1 - μ2)**2 ) / (2*σ2**2)
    C = -1/2
    y = A + B + C
    return y

# KL divergence
def KLdivergence(p,q,dx):
    KL=np.sum(p * np.log(p/q)) * dx
    return KL

#Notch x
dx  = 0.01

#Range of x
xlm = [-6,6]

#x coordinate
x   = np.arange(xlm[0],xlm[1]+dx,dx)

#Number of x
x_n   = len(x)

# Case 1
# p(x) = N(0,1)
# q(x) = N(μ,1)

# p(x)Average μ1
μ1   = 0
# p(x)Standard deviation σ1
σ1   = 1  

# p(x)
px   = gaussian1d(x,μ1,σ1)

# q(x)Standard deviation σ2
σ2   = 1

# q(x)Average μ2
U2   = np.arange(-4,5,1)

U2_n = len(U2)

# q(x)
Qx   = np.zeros([x_n,U2_n])

#KL divergence
KL_U2  = np.zeros(U2_n)

for i,μ2 in enumerate(U2):
    qx        = gaussian1d(x,μ2,σ2)
    Qx[:,i]   = qx
    KL_U2[i]  = KLdivergence(px,qx,dx)


#Scope of analytical solution
U2_exc    = np.arange(-4,4.1,0.1)

#Analytical solution
KL_U2_exc = gaussian1d_KLdivergence(μ1,σ1,U2_exc,σ2)

#Analytical solution 2
KL_U2_exc2 = U2_exc**2 / 2

#
# plot
#

# figure
fig = plt.figure(figsize=(8,4))
#Default color
clr=plt.rcParams['axes.prop_cycle'].by_key()['color']

# axis 1 
#-----------------------
#Normal distribution plot
ax = plt.subplot(1,2,1)
# p(x)
plt.plot(x,px,label='$p(x)$')       
# q(x)
line,=plt.plot(x,Qx[:,i],color=clr[1],label='$q(x)$')       
#Usage Guide
plt.legend(loc=1,prop={'size': 13})

plt.xticks(np.arange(xlm[0],xlm[1]+1,2))
plt.xlabel('$x$')

# axis 2
#-----------------------
#KL divergence
ax2 = plt.subplot(1,2,2)
#Analytical solution
plt.plot(U2_exc,KL_U2_exc,label='Analytical')
#Calculation
point, = ax2.plot([],'o',label='Numerical')

#Usage Guide
# plt.legend(loc=1,prop={'size': 15})

plt.xlim([U2[0],U2[-1]])
plt.xlabel('$\mu$')
plt.ylabel('$KL(p||q)$')

plt.tight_layout()

#Common settings for axes
for a in [ax,ax2]:
    plt.axes(a)
    plt.grid()
    #In a square
    plt.gca().set_aspect(1/plt.gca().get_data_ratio())

#update
def update(i):
    #line
    line.set_data(x,Qx[:,i])
    #point
    point.set_data(U2[i],KL_U2[i])

    #title
    ax.set_title("$\mu_2=%.1f$" % U2[i],fontsize=15)
    ax2.set_title('$KL(p||q)=%.1f$' % KL_U2[i],fontsize=15)

#animation
ani = animation.FuncAnimation(fig, update, interval=1000,frames=U2_n)
# plt.show()
# ani.save("KL_μ.gif", writer="imagemagick")

When the standard deviation is a variable

Next, let 0 be the mean $ \ mu_2 $ of $ q (x) $, and make only the standard deviation $ \ sigma_2 $ a variable.

q(x) =N(0,\sigma^2_2)= \frac{1}{\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{x^2}{2\sigma_2^2}\right)

The KL divergence at this time is

\begin{eqnarray}

KL(p||q) &=& \ln\left(\frac{\sigma_2}{\sigma_1}\right) + \frac{\sigma_1^2+(\mu_1-\mu_2)^2}{2\sigma_2^2} - \frac{1}{2} \\
         &=& \ln\left(\frac{\sigma_2}{1}\right) + \frac{1^2}{2\sigma_2^2} - \frac{1}{2} \\
         &=& \ln\left(\sigma_2\right) + \frac{1}{2\sigma_2^2} - \frac{1}{2} \\
\end{eqnarray}

It will be. \sigma_2Value of 0.Probability distribution when changing from 5 to 4$q(x)And KL divergenceKL(p||q)$The value of is as follows.

KL_σ.gif

As before, the change in KL divergence became 0 when the probability distributions matched, and increased as the shape changed.

import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation

#normal distribution
def gaussian1d(x,μ,σ):
    y = 1 / ( np.sqrt(2*np.pi* σ**2 ) )  * np.exp( - ( x - μ )**2  / ( 2 * σ ** 2 ) )
    return y

#Normal distribution KL divergence
def gaussian1d_KLdivergence(μ1,σ1,μ2,σ2):
    A = np.log(σ2/σ1)
    B = ( σ1**2 + (μ1 - μ2)**2 ) / (2*σ2**2)
    C = -1/2
    y = A + B + C
    return y

# KL divergence
def KLdivergence(p,q,dx):
    KL=np.sum(p * np.log(p/q)) * dx
    return KL

#Notch x
dx  = 0.01

#Range of x
xlm = [-6,6]

#x coordinate
x   = np.arange(xlm[0],xlm[1]+dx,dx)

#Number of x
x_n   = len(x)

# Case 2
# p(x) = N(0,1)
# q(x) = N(0,σ**2)

# p(x)Average μ1
μ1   = 0
# p(x)Standard deviation σ1
σ1   = 1  

# p(x)
px   = gaussian1d(x,μ1,σ1)

# q(x)Average μ2
μ2   = 0

# q(x)Standard deviation σ2
S2   = np.hstack([ np.arange(0.5,1,0.1),np.arange(1,2,0.2),np.arange(2,4.5,0.5) ])

S2_n = len(S2)

# q(x)
Qx   = np.zeros([x_n,S2_n])

#KL divergence
KL_S2  = np.zeros(S2_n)

for i,σ2 in enumerate(S2):
    qx        = gaussian1d(x,μ2,σ2)
    Qx[:,i]   = qx
    KL_S2[i]  = KLdivergence(px,qx,dx)


#Scope of analytical solution
S2_exc    = np.arange(0.5,4+0.05,0.05)

#Analytical solution
KL_S2_exc = gaussian1d_KLdivergence(μ1,σ1,μ2,S2_exc)

#Analytical solution 2
KL_S2_exc2 = np.log(S2_exc) + 1/(2*S2_exc**2) - 1 / 2

#
# plot
#

# figure
fig = plt.figure(figsize=(8,4))
#Default color
clr=plt.rcParams['axes.prop_cycle'].by_key()['color']

# axis 1 
#-----------------------
#Normal distribution plot
ax = plt.subplot(1,2,1)
# p(x)
plt.plot(x,px,label='$p(x)$')       
# q(x)
line,=plt.plot(x,Qx[:,i],color=clr[1],label='$q(x)$')       
#Usage Guide
plt.legend(loc=1,prop={'size': 13})

plt.ylim([0,0.8])
plt.xticks(np.arange(xlm[0],xlm[1]+1,2))
plt.xlabel('$x$')

# axis 2
#-----------------------
#KL divergence
ax2 = plt.subplot(1,2,2)
#Analytical solution
plt.plot(S2_exc,KL_S2_exc,label='Analytical')
#Calculation
point, = ax2.plot([],'o',label='Numerical')

#Usage Guide
# plt.legend(loc=1,prop={'size': 15})

plt.xlim([S2[0],S2[-1]])
plt.xlabel('$\sigma$')
plt.ylabel('$KL(p||q)$')

plt.tight_layout()

#Common settings for axes
for a in [ax,ax2]:
    plt.axes(a)
    plt.grid()
    #In a square
    plt.gca().set_aspect(1/plt.gca().get_data_ratio())

#update
def update(i):
    #line
    line.set_data(x,Qx[:,i])
    #point
    point.set_data(S2[i],KL_S2[i])

    #title
    ax.set_title("$\sigma_2=%.1f$" % S2[i],fontsize=15)
    ax2.set_title('$KL(p||q)=%.1f$' % KL_S2[i],fontsize=15)

#animation
ani = animation.FuncAnimation(fig, update, interval=1000,frames=S2_n)
plt.show()
# ani.save("KL_σ.gif", writer="imagemagick")

When the mean and standard deviation are variables

Below is a plot of the KL divergence values when both the mean $ \ mu_2 $ and the standard deviation $ \ sigma_2 $ are changed.

## bonus ![KL_motion2.gif](https://qiita-image-store.s3.amazonaws.com/0/35426/9b0197d8-1fd6-de81-673f-a8be1be727a4.gif)
import numpy as np
import matplotlib.pyplot as plt

#normal distribution
def gaussian1d(x,μ,σ):
    y = 1 / ( np.sqrt(2*np.pi* σ**2 ) )  * np.exp( - ( x - μ )**2  / ( 2 * σ ** 2 ) )
    return y

#Normal distribution KL divergence
def gaussian1d_KLdivergence(μ1,σ1,μ2,σ2):
    A = np.log(σ2/σ1)
    B = ( σ1**2 + (μ1 - μ2)**2 ) / (2*σ2**2)
    C = -1/2
    y = A + B + C
    return y

# KL divergence
def KLdivergence(p,q,dx):
    KL=np.sum(p * np.log(p/q)) * dx
    return KL

def Motion(event):
    global cx,cy,cxid,cyid
    
    xp = event.xdata
    yp = event.ydata

    if (xp is not None) and (yp is not None):
        gca = event.inaxes

        if gca is axs[0]:
            cxid,cx = find_nearest(x,xp)
            cyid,cy = find_nearest(y,yp)

            lns[0].set_data(G_x,Qx[:,cxid,cyid])
            lns[1].set_data(x,Z[:,cyid])
            lns[2].set_data(y,Z[cxid,:])            
            

            lnhs[0].set_ydata([cy,cy])
            lnvs[0].set_xdata([cx,cx])

            lnvs[1].set_xdata([cx,cx])
            lnvs[2].set_xdata([cy,cy])
            

        if gca is axs[2]:    
            cxid,cx = find_nearest(x,xp)

            lns[0].set_data(G_x,Qx[:,cxid,cyid])
            lns[2].set_data(y,Z[cxid,:])            
            lnvs[0].set_xdata([cx,cx])
            lnvs[1].set_xdata([cx,cx])

        if gca is axs[3]:    
            cyid,cy = find_nearest(y,xp)

            lns[0].set_data(G_x,Qx[:,cxid,cyid])
            lns[1].set_data(x,Z[:,cyid])
            lnhs[0].set_ydata([cy,cy])
            lnvs[2].set_xdata([cy,cy])
            
    axs[1].set_title("$\mu_2=%5.2f, \sigma_2=$%5.2f" % (cx,cy),fontsize=15)
    axs[0].set_title('$KL(p||q)=$%.3f' % Z[cxid,cyid],fontsize=15)

    plt.draw()

def find_nearest(array, values):
    id = np.abs(array-values).argmin()
    return id,array[id]

#Notch x
G_dx  = 0.01
#Range of x
G_xlm = [-4,4]
#x coordinate
G_x   = np.arange(G_xlm[0],G_xlm[1]+G_dx,G_dx)
#Number of x
G_n   = len(G_x)

# p(x)Average μ1
μ1   = 0
# p(x)Standard deviation σ1
σ1   = 1  
# p(x)
px   = gaussian1d(G_x,μ1,σ1)

# q(x)Average μ2
μ_lim = [-2,2]
μ_dx  = 0.1
μ_x   = np.arange(μ_lim[0],μ_lim[1]+μ_dx,μ_dx)
μ_n   = len(μ_x)

# q(x)Standard deviation σ2
σ_lim = [0.5,4]
σ_dx  = 0.05
σ_x   = np.arange(σ_lim[0],σ_lim[1]+σ_dx,σ_dx)
σ_n   = len(σ_x)

#KL divergence
KL   = np.zeros([μ_n,σ_n])
# q(x)
Qx   = np.zeros([G_n,μ_n,σ_n])

for i,μ2 in enumerate(μ_x):
    for j,σ2 in enumerate(σ_x):
        KL[i,j]   = gaussian1d_KLdivergence(μ1,σ1,μ2,σ2)
        Qx[:,i,j] = gaussian1d(G_x,μ2,σ2)

x   = μ_x
y   = σ_x

X,Y = np.meshgrid(x,y)
Z   = KL

cxid  = 0
cyid  = 0

cx    = x[cxid]
cy    = y[cyid]

xlm   = [ x[0], x[-1] ]
ylm   = [ y[0], y[-1] ]

axs   = []
ims   = []
lns   = []
lnvs  = []
lnhs  = []

# figure
#----------------
plt.close('all')
plt.figure(figsize=(8,8))
#Default color
clr=plt.rcParams['axes.prop_cycle'].by_key()['color']

#font size
plt.rcParams["font.size"] = 16
#Line width
plt.rcParams['lines.linewidth'] = 2
#Make grid linestyle a dotted line
plt.rcParams["grid.linestyle"] = '--'

#Eliminate range margins when plotting
plt.rcParams['axes.xmargin'] = 0.

# ax1
#----------------
ax = plt.subplot(2,2,1)

Interval = np.arange(0,8,0.1)
plt.plot(μ1,σ1,'rx',label='$(μ_1,σ_1)=(0,1)$')
im = plt.contourf(X,Y,Z.T,Interval,cmap='hot')
lnv= plt.axvline(x=cx,color='w',linestyle='--',linewidth=1)
lnh= plt.axhline(y=cy,color='w',linestyle='--',linewidth=1)

ax.set_title('$KL(p||q)=$%.3f' % Z[cxid,cyid],fontsize=15)
plt.xlabel('μ')
plt.ylabel('σ')

axs.append(ax)
lnhs.append(lnh)
lnvs.append(lnv)
ims.append(im)

# ax2
#----------------
ax = plt.subplot(2,2,2)
plt.plot(G_x,px,label='$p(x)$')
ln, = plt.plot(G_x,Qx[:,cxid,cyid],color=clr[1],label='$q(x)$')
plt.legend(prop={'size': 10})
ax.set_title("$\mu_2=%5.2f, \sigma_2=$%5.2f" % (cx,cy),fontsize=15)

axs.append(ax)
lns.append(ln)
plt.grid()

# ax3
#----------------
ax = plt.subplot(2,2,3)
ln,=plt.plot(x,Z[:,cyid])
lnv= plt.axvline(x=cx,color='k',linestyle='--',linewidth=1)

plt.ylim([0,np.max(Z)])
plt.grid()
plt.xlabel('μ')
plt.ylabel('KL(p||q)')

lnvs.append(lnv)
axs.append(ax)
lns.append(ln)

# ax4
#----------------
ax = plt.subplot(2,2,4)
ln,=plt.plot(y,Z[cxid,:])

lnv= plt.axvline(x=cy,color='k',linestyle='--',linewidth=1)

plt.ylim([0,np.max(Z)])
plt.xlim([ylm[0],ylm[1]])
plt.grid()

plt.xlabel('σ')
plt.ylabel('KL(p||q)')

lnvs.append(lnv)
axs.append(ax)
lns.append(ln)

plt.tight_layout()

for ax in axs:
    plt.axes(ax)
    ax.set_aspect(1/ax.get_data_ratio())
    
plt.connect('motion_notify_event', Motion)

plt.show()

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