Families of Discrete Random Variables

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INDEX

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Bernoulli Random Variable

Geometric Random Variable

Binomial Random Variable

Negative Binomial Random Variable (Pascal Random Variable)

Discrete Uniform Random Variable

Poisson Random Variable

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Bernoulli Random Variable

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\(X\) is a \(Bernoulli(p)\) random variable if the PMF of \(X\) h 

$$P_{X}(x) = \begin{cases} 1-p & \quad x = 0\\ p & \quad x = 1\\ 0 & \quad \text{otherwise}\\ \end{cases} $$

where the parameter \(p\) is in the range \(0 < p < 1\).

$$\text{Expected value} \quad E[X] = p.$$

$$\text{Second moment} \quad E[X^2] = p.$$

$$\text{Variance} \quad Var[X] = p(1-p).$$

ex) 성공할 확률은 \(p\), 실패할 확률은 \(1 - p\), \(X = \) 성공 또는 실패.

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Geometric Random Variable

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\(X\) is a \(Geomatric(p)\) random variable if the PMF of \(X\) has the form

$$P_{X}(x) = \begin{cases} p(1-p)^{x-1} & \quad x = 1, 2, ...\\ 0 & \quad \text{otherwise.}\\ \end{cases} $$

where the parameter \(p\) is in the range \(0 < p < 1\).

$$\text{Expected value} \quad E[X] = 1/p.$$

$$\text{Second moment} \quad E[X^2] = (2-p)/p.$$

$$\text{Variance} \quad Var[X] = (1-p)/p^2.$$

ex) 성공할 확률은 \(p\), 실패할 확률은 \(1 - p\), \(X = \) 성공할때까지 시행한 횟수(첫 번째 성공까지 걸린 횟수), \(P_X(x) = \) x번째에 처음으로 성공할 확률. \(E[X] = \) 성공할때까지 시행한 횟수의 기댓값. 각 시행은 독립임.

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Binomial Random Variable

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\(X\) is a \(Binomial(n, p)\) random variable if the PMF of \(X\) has the form

$$P_{X}(x) = \binom{n}{x}p^x(1-p)^{n-x}$$

where \(0 < p < 1\) and \(n\) is an integer such that \(n \geq 1\).

$$\text{Expected value} \quad E[X] = np.$$

$$\text{Second moment} \quad E[X^2] = np(np-p+1).$$

$$\text{Variance} \quad Var[X] = np(1-p).$$

ex) 성공할 확률은 \(p\), 실패할 확률은 \(1 - p\), \(X = \) 성공한 횟수, \(n = \) 시도한 횟수, \(P_X(x) = \) n번 시행했을때 x번 성공할 확률. \(E[X] = \) n번 시행했을때, 성공한 횟수의 기댓값. 각 시행은 독립임.

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Negative Binomial Random Variable (Pascal Random Variable)

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\(X\) is a \(NB(r; p)\) random variable if the PMF of \(X\) has the form

$$f(k;r,p)\equiv \Pr(X=k)={\binom {k+r-1}{k}}p^{k}(1-p)^{r}\quad {\text{for }}k=0,1,2,\dotsc$$

where \(k\) is number of successes, \(r\) is number of failures, and \(p\) is the probability of success.

$$\text{Expected value} \quad E[X] = pr/(1-p).$$

$$\text{Second moment} \quad E[X^2] = pr(pr+1)/(1-p)^2.$$

$$\text{Variance} \quad Var[X] = pr/(1-p)^2.$$

$$\text{Skewness (Third normalised moment)} \quad E[X^2] = \frac{1+p}{\sqrt{pr}}$$

ex) 성공할 확률은 \(p\), 실패할 확률은 \(1 - p\), \(k = \) 성공한 횟수, \(r = \) 실패한 횟수, \(P_X(x) = \) r+k번 시행했고, 마지막 시행이 실패면서 k번 성공할 확률.

위의 Negative Binomial Random Variable을 마지막 시행을 성공했을 경우로 바꾸고, 시행횟수를 \(x\)로 정하면, 다음과 같은 PMF를 만들 수 있다. 1번의 성공을 가정했기 때문에 k는 1보다 크거나 같아야 한다.

\(X\) is a \(Pascal(k, p)\) random variable if the PMF of \(X\) has the form

$$P_{X}(x) = \binom{x-1}{k-1}p^k(1-p)^{x-k}$$

where \(0 < p < 1\) and \(k\) is an integer such that \(k \geq 1\).

$$\text{Expected value} \quad E[X] = k/p.$$

$$\text{Second moment} \quad E[X^2] = k(1-p)+k^2/p^2.$$

$$\text{Variance} \quad Var[X] = k(1-p)/p^2.$$

ex) 성공할 확률은 \(p\), 실패할 확률은 \(1 - p\), \(k = \) 성공한 횟수, \(x = \) 시행한 횟수, \(P_X(x) = \) x번 시행했고, 마지막 시행이 성공이면서 k번 성공 했을 확률.

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Discrete Uniform Random Variable

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\(X\) is a \(discrete uniform (k, l)\) random variable if the PMF of \(X\) has the form

$$P_{X}(x) = \begin{cases} l/(l-k+1) & \quad x = k, k +1, k+2, \dotsc , l\\ 0 & \quad otherwise\\ \end{cases} $$

where the parameters \(k\) and \(l\) are integers such that \(k < l\).

$$\text{Expected value} \quad E[X] = (l + k)/2$$

$$\text{Variance} \quad Var[X] = (l - k)(l - k +2)/12.$$

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Poisson Random Variable

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\(X\) is a \(Poisson(\alpha)\) random variable if the PMF of \(X\) has the form

$$P_{X}(x) = \begin{cases} \alpha^xe^{-\alpha}/x! & \quad x = 0, 1, 2, \dotsc, \\ 0 & \quad \text{otherwise}\\ \end{cases} $$

where the parameter \(\alpha\) is in the range \(\alpha > 0\).

$$\text{Expected value} \quad E[X] = \alpha$$

$$\text{Variance} \quad Var[X] = \alpha$$

$$\text{Skewness (Third normalised moment)} \quad E[X^2] = \alpha^{-1/2}$$

ex) \(\alpha = \lambda T = \) T시간동안 이벤트가 발생한 횟수,\(\lambda = \) T의 단위에 대한 평균 이벤트 발생율.(T가 시간일 경우 단위 시간당 평균 이벤트 발생 횟수). \(T = \) 관찰한 시간, \(P_X(x) = \) 관찰 시간 T 동안 j번 이벤트가 발생할 확률.

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Proof - Bernoulli

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PMF of \(Bernoulli(p)\) is

$$P_{X}(x) = \begin{cases} 1-p & \quad x = 0\\ p & \quad x = 1\\ 0 & \quad \text{otherwise}\\ \end{cases} $$

Expected Value

\( E[X] = \mu_X = \displaystyle\sum_{x \in S_X} x P_X(x)\\ E[X] = 0 \times (1 - p) + 1 \times p = p\\ \quad \\ \therefore E[X] = p \)

Variance

\( Var[X] = E[X^2] - (E[X])^2\\ E[X^2] = 0^2 \times (1 - p) + 1^2 \times p = p\\ Var[X] = p - p^2\\ \quad \\ \therefore Var[X] = p(1 - p) \)

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Proof - Geometric

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PMF of \(Geometric(p)\) is

$$P_{X}(x) = \begin{cases} p(1-p)^{x-1} & \quad x = 1, 2, ...\\ 0 & \quad \text{otherwise.}\\ \end{cases} $$

Expected Value

\( \begin{align} E[X] &= \displaystyle\sum_{x = 1}^{\infty} x p(1-p)^{x - 1}\\ &= p + 2p(1-p) + 3p(1-p)^2 + \dots\\ \end{align} \)

\( \begin{align} (1-p)E[X] &= \displaystyle\sum_{x = 1}^{\infty} x p(1-p)^{x}\\ &= p(1-p) + 2p(1-p)^2 + 3p(1-p)^3 + \dots\\ \end{align} \)

\( \begin{align} E[X] - (1-p)E[X] &= p E[X] = p + p(1 - p) + p(1 - p)^2 + \dots\\ &= \displaystyle\sum_{x = 1}^{\infty} p(1 - p)^{x - 1} = \frac{p}{1 - (1 - p)} = 1\\ \end{align} \)

\(\therefore E[X] = \frac{1}{p}\)

Variance

\( \begin{align} E[X^2] &= \displaystyle\sum_{x = 1}^{\infty} x^2 p(1-p)^{x - 1}\\ &= p + 4p(1-p) + 9p(1-p)^2 + \dots\\ \end{align} \)

\( \begin{align} E[X^2] - (1-p)E[X^2] &= p E[X^2]\\ &= 1 + 2(1-p) + 2(1-p)^2 + 2(1-p)^3 + \dots\\ &= 1 + \frac{2(1 - p)}{1 - (1 - p)} = \frac{2 - p}{p} \quad \because 1 - p < 1 \\ \end{align} \)

\( E[X^2] = \frac{2 - p}{p^2} \)

\( \begin{align} \therefore Var[X] &= E[X^2] - (E[X])^2\\ &=\frac{2-p}{p^2} - \frac{1}{p^2}\\ &=\frac{1-p}{p^2} \end{align} \)

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Proof - Binnomial

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PMF of \(Binomial(n, p)\) is

$$P_{X}(x) = \binom{n}{x}p^x(1-p)^{n-x}$$

Expected Value

\( \begin{align} E[X] &= \displaystyle\sum_{x = 0}^{n} x p^x(1 - p)^{n-x}\\ &= \displaystyle\sum_{x = 0}^{n} \frac{n!}{(n-x)!x!}x \cdot p^x (1 - p)^{n-x}\\ &= \displaystyle\sum_{x = 1}^{n} \frac{n!}{(n-x)!(x - 1)!} p^x (1 - p)^{n-x}\quad \quad \because \binom{n}{0} 0 \times p^0 (1 - p)^{n}\\ &= \displaystyle\sum_{x = 1}^{n} \frac{n (n - 1)!}{(n-x)!(x - 1)!} p \cdot p^{x - 1} (1 - p)^{n-x}\\ &= np\displaystyle\sum_{x = 1}^{n} \frac{(n - 1)!}{(n-x)!(x - 1)!} p^{x - 1} (1 - p)^{n-x}\\ \\ &\text{Let } a = x - 1, b = n - 1 \text{ then},\\ \\ &= np\displaystyle\sum_{a = 0}^{n} \frac{b!}{(b - a)!a!} p^{a} (1 - p)^{b - a}\\ &= np \quad \quad \because \displaystyle\sum_{a = 0}^{n} \frac{b!}{(b - a)!a!} p^{a} (1 - p)^{b - a} = 1\\ \end{align} \)

\( \begin{align} \therefore E[X] = np\\ \end{align} \)

Variance

\( \begin{align} E[X(X - 1)] &= E[X^2] - \mu = E[X^2] - \mu^2 + \mu^2 - \mu\\ &= Var[X] + \mu^2 - \mu \end{align} \)

\( Var[X] = E[X(X - 1)] - \mu^2 + \mu\\ \)

\( \begin{align} E[X(X - 1)] &= \displaystyle\sum_{x = 0}^{n} x (x - 1) \binom{n}{x}p^{x}(1 - p)^{n - x}\\ &= \displaystyle\sum_{x = 2}^{n} x (x - 1) \binom{n}{x}p^{x}(1 - p)^{n - x} \quad \quad \because x (x - 1) \binom{n}{x}p^{x}(1 - p)^{n - x} = 0 \quad \text{ when } x = 0, 1\\ &= \displaystyle\sum_{x = 2}^{n} x (x - 1) \frac{n!}{(n-x)!x!} p^{x} (1 - p)^{n-x}\\ &= n(n - 1) p^2\displaystyle\sum_{x = 2}^{n} x (x - 1) \frac{(n - 2)!}{(n-x)!x!} p^{x - 2} (1 - p)^{n-x}\\ &= n(n - 1) p^2\displaystyle\sum_{x = 2}^{n} \frac{(n - 2)!}{(n-x)!(x - 2)!} p^{x - 2} (1 - p)^{n-x}\\ \\ &\text{Let } a = x - 2, b = n - 2 \text{ then},\\ \\ &= n(n - 1) p^2\displaystyle\sum_{a = 0}^{b} \frac{b!}{(b - a)!a!} p^{a} (1 - p)^{b - a}\\ &= n(n - 1) p^2 \quad \quad \because \displaystyle\sum_{a = 0}^{b} \frac{b!}{(b - a)!a!} p^{a} (1 - p)^{b - a} = 1\\ \end{align} \)

\( \begin{align} \therefore Var[X] &= n(n-1)p^2 - n^2p^2 + np = np - np^2\\ &= np(1 - p) \end{align} \)