Discrete Random Variable
: There are only TWO possible outcomes. (e.g. male or female, success or failure, 1 or 0)
\bigstar f(x)=p^x(1-p)^{1-x} , x=0 or 1 (two outcomes),
0\leq p\leq 1 (probability is always between 0 and 1)
\bigstar E(X)=p, Var(x)=p(1-p)=pq (where q=1-p)
Proof
f(1|p)=p^1(1-p)^{1-1}=p (probability of being 1)
f(0|p)=p^0(1-p)^1=1-p (probability of being 0)
If we have n data (n samples), we can calculate the likelihood by using joint distribution.
\Rightarrow P(X_{1}=x_{1},X_{2}=x_{2},...,X_{n}=x_{n})=L=\prod_{i=1}^{n}p^{x_{i}}(1-p)^{1-x_{i}}
To maximize? Take a derivative with respect to P and set the equation to equal to 0.
\Rightarrow P=\frac{dL}{dP}=0 \Rightarrow \hat{p}_{MLE}
(we need a chain rule b/c consisting of product)
\Rightarrow log L=l=log(\prod_{i=1}^{n}p^{x_{i}}(1-p)^{1-x_{i}}=\sum_{i=1}^{n} log(p^{X_i}(1-p)^{1-X_{i}})
=\sum_{i=1}^{n}[{x_{i}\cdot logp+(1-X_{i}})\cdot log(1-p) ]
= logp \sum_{i=1}^{n}X_{i}+log(1-p)\sum_{i=1}^{n}(1-X_{i})
= n\bar{X}\cdot log\hat{p}+n(1-\bar{X})\cdot log(1-\hat{p})
\because \sum_{i=1}^{n}X_{i}=n\bar{X}\Rightarrow \bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}
\Rightarrow \frac{dl}{d\hat{p}}=\frac{n\bar{X}}{\hat{p}} - \frac{n(1-\bar{X})}{1-\hat{p}} = 0 (for maximizing)
= \frac{n\bar{X}}{\hat{p}}=\frac {n(1-\bar{X})}{1-\hat{p}}=\bar{X}\cdot \hat{p}\bar{X}=\hat{p}-\hat{p}\bar{X} \Rightarrow \hat{p}_{MLE}=\bar{X}
\bigstar Show that T=\sum_{i=1}^{n}X_{i} is a sufficient statistic.
Proof
By independence, the joint distribution of the random sample is
\prod_{i}^{n}p^{x_{i}}(1-p)^{1-x_{i}}=p^{\sum X_{i}}(1-p)^{n-\sum X_{i}} \cdot 1 ,
where p^{\sum X_{i}}(1-p)^{n-\sum X_{i}} = g(\sum x_{i},p) , and 1= h(X_{1},...,X_{n})
\bigstar Show that Bernoulli distribution is part of the exponential family.
Proof
We need to show f_{\theta}(X)= \exp {[\sum_{i=1}^{k}]C_{i}(\theta)\cdot T_{j}(X)}+d(\theta)+ s(X) Click link to more details
parameter p, where p= P(X=1)
p(x|p)=p^x(1-p)^{1-x}
p(x|p)=exp{[log(p^x(1-p)^{1-x})]}=exp[x\cdot logp+(1-x)\cdot log(1-p)]
= exp [x\cdot log \frac{p}{1-p} + log(1-p)]
This shows the Bernoulli distribution belongs to the exponential family with parameter c(\theta)=\log \frac {p}{1-p}, T(x)=x, d(\theta)=\log (1-p), s(x)=0
Likelihood Ratio Example
\bigstar Show that T=\sum_{i=1}^{n}X_{i} is a sufficient statistic.
Proof
By independence, the joint distribution of the random sample is
\prod_{i}^{n}p^{x_{i}}(1-p)^{1-x_{i}}=p^{\sum X_{i}}(1-p)^{n-\sum X_{i}} \cdot 1 ,
where p^{\sum X_{i}}(1-p)^{n-\sum X_{i}} = g(\sum x_{i},p) , and 1= h(X_{1},...,X_{n})
\bigstar Show that Bernoulli distribution is part of the exponential family.
Proof
We need to show f_{\theta}(X)= \exp {[\sum_{i=1}^{k}]C_{i}(\theta)\cdot T_{j}(X)}+d(\theta)+ s(X) Click link to more details
parameter p, where p= P(X=1)
p(x|p)=p^x(1-p)^{1-x}
p(x|p)=exp{[log(p^x(1-p)^{1-x})]}=exp[x\cdot logp+(1-x)\cdot log(1-p)]
= exp [x\cdot log \frac{p}{1-p} + log(1-p)]
This shows the Bernoulli distribution belongs to the exponential family with parameter c(\theta)=\log \frac {p}{1-p}, T(x)=x, d(\theta)=\log (1-p), s(x)=0
Likelihood Ratio Example
Y_{1},...,Y_{n} denote a random sample from Bernoulli P(Y_{i}|p)=p^{y_{i}}(1-p)^{1-y_{i}} , where y_{i}=0 or 1. Suppose H_{0}:P=P_{0} , H_{1}:P=P_{a}, where P_{0} < P_{a}
(a) Show that \frac{L(P_{o})}{L(P_{a})}=[\frac{P_{0}\cdot (1-P_{a})}{(1-P_{0}) \cdot P_{a}}]^{\sum y_{i}}\cdot (\frac{1-P_{0}}{1-P_{a}})^n
(b) Argue that \frac{L(P_{o})}{L(P_{a})} < K iff \sum y_{i} < k
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