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"### Machine Learning\n",
"#### Maximum Likelihood Estimation\n",
"Given a number of samples, say $n$, received from the environment, and knowing their joint probability function denoted by $f_n(\\boldsymbol{x};\\boldsymbol{w})$, we can define the **likelihood function** as shown below:\n",
"<br>$L(\\boldsymbol{x},\\boldsymbol{w})=f_n(\\boldsymbol{x};\\boldsymbol{w})$ (1)\n",
"<br>where $\\boldsymbol{w}$ is generally a vector of unknown parameters of the joint probability function $f_n(.;.)$.\n",
"<br>Assuming the samples are *independent and identically distributed* (i.i.d.) random variables, we can simplify Eq. (1) to this:\n",
"<br>$L(\\boldsymbol{x};\\boldsymbol{w})=\\prod_{i=1}^{n}f(x_i;\\boldsymbol{w})$ (2)\n",
"<br>where $f(x_i;\\boldsymbol{w})$ is the probability function of $x_i$ with parameter vector $\\boldsymbol{w}$ \n",
"<br>The **Maximum likelihood estimation** finds the optimal $\\boldsymbol{w}$ from:\n",
"<br>$\\boldsymbol{w}=argmax_\\boldsymbol{w} L(\\boldsymbol{x};\\boldsymbol{w})$ (3)\n",
"<br>If the likelihood function is *differentiable*, sufficient conditions for the **MLE** are:\n",
"<br>$\\frac{\\partial L}{\\partial w_0}=0$, $\\frac{\\partial L}{\\partial w_1}=0$,...,$\\frac{\\partial L}{\\partial w_{q-1}}=0$ (4)\n",
"<br><br>**Contents:**\n",
"- Estimation by the MLE for samples from a **continuous uniform distribution**.\n",
"- Estimation by the MLE for samples from a **normal distribution**.\n",
"<hr>\n",
"https://github.com/ostad-ai/Machine-Learning\n",
"<br> Explanation: https://www.pinterest.com/HamedShahHosseini/Machine-Learning"
]
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"cell_type": "code",
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"source": [
"# importing the required modules\n",
"import numpy as np"
]
},
{
"cell_type": "markdown",
"id": "ccd9bb5d",
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"It can be shown that for samples of **i.i.d.** random variables taken from a **uniform distribution**,\n",
"say $f(x_i,;w_0,w_1)$, the **MLE** leads to the following parameter estimation:\n",
"<br> $w_0=min(x_1,x_2,...,x_n)$ and $w_1=max(x_1,x_2,...,x_n)$\n",
"<br> In the following example, we investigate these formulae:"
]
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{
"cell_type": "code",
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"name": "stdout",
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"We get 100 samples from a uniform distribution\n",
"The truth values of parameters: a=3 and b=10\n",
"The estimated parameters by MLE: a=3.016066812589626,b=9.723057963740676\n"
]
}
],
"source": [
"# estimation of parameters for continuous uniform distribution by the MLE\n",
"def MLE_uniform(samples):\n",
" w0=np.min(samples)\n",
" w1=np.max(samples)\n",
" return w0,w1\n",
"\n",
"#---example\n",
"a,b,n=3,10,100\n",
"samples=a+(b-a)*np.random.random(120)\n",
"w0,w1=MLE_uniform(samples)\n",
"print(f'We get {n} samples from a uniform distribution')\n",
"print(f'The truth values of parameters: a={a} and b={b}')\n",
"print(f'The estimated parameters by MLE: a={w0},b={w1}')"
]
},
{
"cell_type": "markdown",
"id": "59830c63",
"metadata": {},
"source": [
"For $n$ i.i.d. samples taken from a normal distribution, we cam estimate the parameters (mean and variance) by:\n",
"<br>$mean=\\frac{1}{n}\\sum_{i=1}^{n}x_i$\n",
"<br>$variance=\\frac{1}{n}\\sum_{i=1}^{n}(x_i-mean)^2$"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "74ae0061",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"We get 1000 samples from a normal distribution\n",
"The truth values of parameters: mean=5 and variance=15\n",
"The estimated parameters by MLE: mean=5.211299243717775,variance=15.08865066577488\n"
]
}
],
"source": [
"# example of samples taken from a normal distribution\n",
"def MLE_normal(samples):\n",
" mean=np.mean(samples)\n",
" variance=np.mean((samples-mean)**2)\n",
" return mean, variance\n",
"m,var,n=5,15,1000\n",
"samples=m+np.sqrt(var)*np.random.randn(n)\n",
"mean,variance=MLE_normal(samples)\n",
"print(f'We get {n} samples from a normal distribution')\n",
"print(f'The truth values of parameters: mean={m} and variance={var}')\n",
"print(f'The estimated parameters by MLE: mean={mean},variance={variance}')"
]
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