Be sure to import Symbulate using the following commands.
from symbulate import *
%matplotlib inline
Introduction to discrete distributions and random variables¶
A discrete probability distribution assigns positive probability to the values in some countable set. A discrete distribution can be specified by a probability mass function (pmf), $p(\cdot)$, which maps a value $x$ to its probability $p(x)\in [0,1]$. A probability mass function must satisfy $\sum_x p(x) = 1$.
A discrete random variable $X$ defined on a probability space with probability measure $P$ has a probability mass function $p_X$ defined by $p_X(x) = P(X = x)$. The expected value of a discrete random variable is
$$ E(X) = \sum_x x\, p_X(x) = \sum_x x P(X=x) $$Defining a general discrete distribution with BoxModel¶
A discrete distribution can be specified via a BoxModel where the tickets represent possible values (i.e. the values with non-zero probability) and the probs option specifies the probabilities.
Example. A certain probability mass function $p$ satisfies $p(1) = 0.4$, $p(2) = 0.2$, $p(5) = 0.3$, $p(10)=0.1$, and $p(x)=0$ otherwise.
P = BoxModel([1, 2, 5, 10], probs=[0.4, 0.2, 0.3, 0.1])
P.sim(10000).tabulate(normalize=True)
RV(P).sim(10000).plot()
Discrete Uniform¶
A discrete Uniform distribution with parameters $a$ and $b$, DiscreteUniform(a, b), assigns equal probability to each integer between $a$ and $b$ (inclusive), as specified by the probability mass function
If $X$ is a random variable with a DiscreteUniform(a, b) distribution then
Example. Actual and simulated pmf for DiscreteUniform(a=1, b=6).
sims = RV(DiscreteUniform(a=1, b=6)).sim(10000)
sims.plot()
DiscreteUniform(a=1, b=6).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(DiscreteUniform(a=1, b=6).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(DiscreteUniform(a=1, b=6).var()))
Example. DiscreteUniform(a = 1, b) pmf.
bs = [3, 5, 7]
for b in bs:
DiscreteUniform(a=1, b=b).plot()
plt.legend(["b=" + str(i) for i in bs])
plt.title("Discrete Uniform distributions with a=1")
Bernoulli distributions¶
A Bernoulli($p$) distribution takes value 1 with probability $p$ and 0 with probability $1-p$.
If $X$ is a random variable with a Bernoulli($p$) distribution then
$$ \begin{align*} E(X) & = p \\ Var(X) & = p(1-p) \end{align*} $$Example. Simulated pmf for Bernoulli(p=0.3).
RV(Bernoulli(0.3)).sim(10000).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Bernoulli(0.3).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Bernoulli(0.3).var()))
Binomial distributions¶
Binomial distributions are often used to model the number of successes observed in a fixed number (n) of Bernoulli(p) trials (independent, success/failure trials, with probability of success p on each trial). A Binomial distribution with parameters n (a nonnegative integer) and p (in [0,1]) is specified by the probability mass function
If $X$ is a random variable with a Binomial($n, p$) distribution then
$$ \begin{align*} E(X) & = np \\ Var(X) & = np(1-p) \end{align*} $$Example. Actual and simulated pmf for Binomial(n=10, p=0.4).
sims = RV(Binomial(10, 0.4)).sim(10000)
sims.plot()
Binomial(10, 0.4).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Binomial(10, 0.4).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Binomial(10, 0.4).var()))
Example. Binomial(n=10, p) pmf.
ps = [0.10, 0.50, 0.75]
for p in ps:
Binomial(10, p).plot()
plt.legend(["p=" + str(i) for i in ps])
plt.title("Binomial distributions with n=10")
Example. Binomial(n, p=0.4) pmf.
ns = [5, 10, 15]
for n in ns:
Binomial(n, 0.4).plot()
plt.legend(["n=" + str(i) for i in ns])
plt.title("Binomial distributions with p=0.4")
Hypergeometric distributions¶
Hypergeometric distributions are used to model the number of "successes" in n (a nonnegative integer) draws without replacement from a box containing $N_0$ failures and $N_1$ successes. A hypergeometric distribution is specified by the probability mass function
$$
\frac{\binom{N_1}{x}\binom{N_0}{n-x}}{\binom{N_0 + N_1}{n}}, \quad x = 0,1, \ldots, n;\; x \le N_1;\; n-x \le N_0
$$
If $X$ is a random variable with a Hypergeometric($n, N_0, N_1$) distribution then
$$
\begin{align*}
E(X) & = \frac{n N_1}{N_0 + N_1} \\
Var(X) & = \frac{n N_1(N_0)(N_0 + N_1 - n)}{(N_0 + N_1)^2(N_0 + N_1 - 1)}
\end{align*}
$$
Example. Actual and simulated pmf for Hypergeometric(n=5, N0=5, N1=10).
sims = RV(Hypergeometric(n=5, N0=5, N1=10)).sim(10000)
sims.plot()
Hypergeometric(n=5, N0=5, N1=10).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Hypergeometric(n=5, N0=5, N1=10).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Hypergeometric(n=5, N0=5, N1=10).var()))
Example. Hypergeometric(n=10, N0=10, N1) pmf.
N1s = [5, 15, 25]
for N1 in N1s:
Hypergeometric(10, 10, N1).plot()
plt.legend(["N1=" + str(i) for i in N1s])
plt.title("Hypergeometric distributions with N0=10, n=10")
Example. Hypergeometric(n=10, N0, N1= 10) pmf.
N0s = [5, 15, 25]
for N0 in N0s:
Hypergeometric(10, N0, 10).plot()
plt.legend(["N0=" + str(i) for i in N0s])
plt.title("Hypergeometric distributions with N1=10, n=10")
Poisson distributions¶
A Poisson distribution with parameter $\lambda>0$ is specified by the probability mass function $$ \frac{e^{-\lambda} \lambda^x}{x!}, \quad x = 0, 1, 2, \ldots $$ If $X$ is a random variable with a Poisson($\lambda$) distribution then $$ \begin{align*} E(X) & = \lambda \\ Var(X) & = \lambda \end{align*} $$
Example. Actual and simulated pmf for Poisson(lam=5).
sims = RV(Poisson(lam = 5)).sim(10000)
sims.plot()
Poisson(lam = 5).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Poisson(lam=5).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Poisson(lam=5).var()))
Example. Poisson($\lambda$) pmf.
lams = [5, 10, 15]
for lam in lams:
Poisson(lam).plot()
plt.legend(["lambda=" + str(i) for i in lams])
plt.title("Poisson distributions with parameter lambda")
Geometric distributions¶
Geometric distributions are often used to model the number of Bernoulli($p$) (independent, success/failure) trials needed to achieve the first success. A geometric distribution with parameter p (in [0,1]) is specified by the probability mass function
$$
p(1-p)^{x-1}, \quad x = 1, 2, 3,\ldots
$$
If $X$ is a random variable with a Geometric(p) distribution then
$$
\begin{align*}
E(X) & = \frac{1}{p} \\
Var(X) & = \frac{(1 - p)}{p^2}
\end{align*}
$$
Example. Actual and simulated pmf for Geometric(p=0.4).
sims = RV(Geometric(p=0.4)).sim(10000)
sims.plot()
Geometric(p=0.4).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Geometric(p=0.4).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Geometric(p=0.4).var()))
Example. Geometric(p) pmf.
ps = [0.25, 0.50, 0.75]
for p in ps:
Geometric(p).plot()
plt.legend(["p=" + str(i) for i in ps])
plt.title("Geometric distributions with parameter p")
Negative binomial distributions¶
Negative binomial distributions are often used to model the number of Bernoulli($p$) (independent, success/failure) trials needed to achieve a specified number of successes ($r$). A negative binomial distribution with parameters $p$ and $r$ is specified by the probability mass function $$ \binom{x-1}{r-1}p^r(1-p)^{x-r}, \quad x = r, r+1, r+2,\ldots $$ If $X$ is a random variable with a NegativeBinomial($r, p$) distribution then $$ \begin{align*} E(X) & = \frac{r}{p} \\ Var(X) & = \frac{r(1 - p)}{p^2} \end{align*} $$
Example. Actual and simulated pmf for NegativeBinomial(r=3, p=0.4).
sims = RV(NegativeBinomial(r=3, p=0.4)).sim(10000)
sims.plot()
NegativeBinomial(r=3, p=0.4).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(NegativeBinomial(r=3, p=0.4).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(NegativeBinomial(r=3, p=0.4).var()))
Example. NegativeBinomial(r, p=0.5) pmf.
rs = [5, 10, 15]
for r in rs:
NegativeBinomial(r, p=0.5).plot()
plt.legend(["r=" + str(i) for i in rs])
plt.title("Negative Binomial distributions with p=0.5")
Example. NegativeBinomial(r=5, p) pmf.
ps = [0.25, 0.50, 0.75]
for p in ps:
NegativeBinomial(r=5, p=p).plot()
plt.legend(["p=" + str(i) for i in ps])
plt.title("Negative Binomial distributions with r=5")
Pascal distributions¶
A Pascal distribution has the same shape as a negative binomial distribution, but the distribution is supported on $\{0, 1, 2, \ldots\}$ rather than $\{r, r+1, r+2, \ldots\}$ with the probability mass function $$ \binom{x+r-1}{r-1}p^r(1-p)^{x}, \quad x = 0, 1, 2,\ldots $$ If Bernoulli($p$) trials are performed until $r$ successes are achieved, then the number of failures observed will have a Pascal distribution.
If $X$ is a random variable with a Pascal($r, p$) distribution then $$ \begin{align*} E(X) & = \frac{r(1 - p)}{p} \\ Var(X) & = \frac{r(1 - p)}{p^2} \end{align*} $$
Example. Actual and simulated pmf for Pascal(r=3, p=0.4).
sims = RV(Pascal(r=3, p=0.4)).sim(10000)
sims.plot()
Pascal(r = 3, p = 0.4).plot()
print('The simulated mean is: {:.2f}'.format(sims.mean()))
print('The true mean is: {:.2f}'.format(Pascal(r=3, p=0.4).mean()))
print('The simulated variance is: {:.2f}'.format(sims.var()))
print('The true variance is: {:.2f}'.format(Pascal(r=3, p=0.4).var()))
Example. Pascal(r, p=0.5) pmf.
rs = [5, 10, 15]
for r in rs:
Pascal(r, p=0.5).plot()
plt.legend(["r=" + str(i) for i in rs])
plt.title("Pascal distributions with p=0.5")
Example. Pascal(r=5, p) pmf.
ps = [0.25, 0.50, 0.75]
for p in ps:
Pascal(r=5, p=p).plot()
plt.legend(["p=" + str(i) for i in ps])
plt.title("Pascal distributions with r=0.5")
