Random Variable

Discrete and Continuous Probability Distributions

Abdullah Al Mahmud

Concepts

Random Variable

Are these correct?

  • Discrete
    • No. of people dying each day
    • Number of heads in successive tosses.
    • Heights of people in Bangladesh
  • Continuous
    • GPA of students
    • Grade of students in individual subjects
    • Income tax paid by people

Integration

  • \(\int x^3+2x\)
  • \(\int_2^3 x^2+x\)
  • Relationship between integration (I) and differentiation(D).
Answer

Probability Distribution

Example

Results of an unbiased die throw

x 1 2 3 4 5 6
P \(\frac 1 6\) \(\frac 1 6\) \(\frac 1 6\) \(\frac 1 6\) \(\frac 1 6\) \(\frac 1 6\)

Biased

x 1 2 3 4 5 6
P \(\frac 1 7\) \(\frac 2 7\) \(\frac 1 7\) \(\frac 1 7\) \(\frac 1 7\) \(\frac 1 7\)

Number of Heads in Coin Toss

A coin is tossed twice

S = {HH, HT, TH, TT}

  • X = number of heads in the tosses
  • What are the possible values of x?
  • x = 0, 1, 2

Fill the probabilities

x 0 1 2
p(x)
  • N:B: X is variable & x is value

PDF vs PMF Curve

PDF

Continuous

PMF

Discrete

Integration and Area

Source: hyperphysics

  • Area = Integration = Probability (if [0,1])
  • \(Area = height \times width = \int f(x) \times dx\)

PDF and PMF Conditions

PDF

Source: upgrad.com

  • \(0 \le f(x) \le 1\)
  • \(\displaystyle \int_{-\infty}^\infty f(x)dx = 1\)

PMF

  • \(0 \le P(X=x) \le 1\)
  • \(\sum P(x_i) = 1\)
  • Thus the sum of all possible outcomes is zero

PDF and PMF Properties

Probability Density Function (continuous)

  • \(\displaystyle \int_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} = 1\)
  • \(P\left( {a \le X \le b} \right) = \int_{{\,a}}^{{\,b}}{{f\left( x \right)\,dx}}\)
  • \(P(X=x)=0\) (theoretically, why?)

Probability Mass Function (discrete)

  • \(\displaystyle \sum_{{\, - \infty }}^{{\,\infty }}{{f\left( x \right)\,dx}} = 1\)
  • \(P_x(X) = P(X=x)\)

PMF Problem 01

\(\displaystyle P(x) = \frac{x+1}{k}; x= 1,2,3,4\)

  • k = ?
  • \(P(X \ge2)\)
  • \(P(X \le 3)\)
  • \(P(2 < X \le 5\)

PMF Problem 02

Given, \(P(x) = \frac{2x+k}{56}; x = -3, -2, -1, 0, 1, 2, 3\)

Discrete or Continuous?

  • k = ?
  • Find probability of each value of x
  • Find \(P(-2 \le x \le 2)\)

More PMF

  • \(P(x) = \frac{1}{14} (a+2x); x = -3, -2, -1, 0, 1, 2, 3\)
  • \(P(x) = k(x-2); x = 3, 4,5,6,7,8\)
  • \(P(x) = \frac{x-1}{k}; x=2,3,4,5\)
  • \(P(x) = \frac{3-|4-x|}{k}; x=2,3,4,5,6\)
  • \(p(x) = \frac{x+4}{30}; x=0,1,2,3,4\)

PMF problem from coin

An unbiased coin is tossed four times and the number of times the heads are obtained is denoted by x. Determine the probability mass function.

  • \(S = \{HHHH, HHHT, \cdots\}\)
  • \(x = 0, 1, 2, 3, 4\)
  • \(P(X = 2) = \frac{4C_2}{2^4}\)
  • \(P(x) = \frac{4C_X}{2^4}\)

PDF Problem 01

\(\displaystyle f(x) = kx(x-1); 1\le x \le 5\)

  • Find k
  • \(P(X>1)\)
  • \(P(X \le 3)\)
  • \(P(2 \le x < 4)\)

PDF Problem 02

\(f(x) = k(x+1); 0\lt x \lt 1\)

  • \(P(X=2)=?\)
  • \(k=?\)
  • \(P(0.4 \lt X \lt 2)=?\)
  • \(\int_{0.4}^1 f(x) + \int_{1}^2 f(x) \to 0\)

More PDF

  • \(f(x) = 2x; 0 < x < 1\)
  • \(f(x) = \frac{1}{30} (3+2x); 2 < x < 5\)
  • \(f(x) = ax^2; 0 < x < 4\)
  • \(f(x) = kx^2 + kx + \frac 18; 0 < x < 8\)
  • \(f(x) = kx; 0 < x < 4\)
  • \(f(x) = 3x^2; 0 \le x \le 1\)
  • \(f(y) = k(3y+5); 1 < y < 5\)
  • \(f(z) = \frac29 (3z-z^2); 0 \le x \le 3\)

Problems Archive

Look here

Cumuluative Distribution Function

F(x) or cdf accumulates all of the probability less than or equal to.

x 1 2 3 4 5 6
P (x) \(\frac 1 7\) \(\frac 1 7\) \(\frac 2 7\) \(\frac 1 7\) \(\frac 1 7\) \(\frac 1 7\)
F (x) \(\frac 1 7\) \(\frac 2 7\) \(\frac 4 7\) \(\frac 5 7\) \(\frac 6 7\) \(1\)

Find

  1. \(P(X<4)\)
  2. \(P(3<X<6)\)

cdf definition

\(F_X(x) = P(X\le x)\)

Discrete

\[F(x) = \sum_{X\le x} P(x)\]

Continuous

  • \[F_{X}(x) = \int_{-\infty}^x f_X(t)dt\]
  • Find cdf for \(f(x) = 2x; 0\le x \le 1\)
  • \(\int \to x^2\)
Answer

cdf properties

  • \(P(a\le x \le b) = F(b)-F(a)\)for \(a\lt b\); what if \(a \lt x \lt b?\)
  • For continuous x, \(f(x) = \frac{d}{dx}[F(x)]\)
  • \(F(-\infty) =0 , F(+\infty) = 1\)

Joint Probability Function

Let, \(I = Infected\), and \(V = Vaccinated\)

\(I\) \(\bar I\) Total
\(V\) 3 276 279
\(\bar V\) 66 473 539
Total 69 749 818

Find the probability that

  1. a vaccinated person is infected
  2. a non-vaccinated person is uninfected
  • These are joint probabilities \(\to\) P(x,y)
  • \(P(x,y) =P(x) \cdot P(y)\) if \(x\) and \(y\) are independent.

Joint-Marginal-Conditional

Let, \(I = Infected\), and \(V = Vaccinated\)

\(I\) \(\bar I\) Total
\(V\) 3 276 279
\(\bar V\) 66 473 539
Total 69 749 818

Find the probabilities that

  • a vaccinated person is infected
  • a non-vaccinated person is uninfected
  • vaccinated if infected
  • infected if not vaccinated
  • vaccinated
  • uninfected

Joint PF Properties

  • \(P(x,y) \ge 0\)
  • \(\Sigma\Sigma P(x,y)=1\)

Coin-Die

Die/Coin 1 2 3 4 5 6
H (1) H1 H2 H3 H4 H5 H6
T (0) T1 T2 T3 T4 T5 T6

X = Outcome of coin toss

Y = Outcome of die throw

x = 0, 1; y = 1, 2, 3, 4, 5, 6

Construct the distribution.

Joint-Marginal-Conditional Revisited

Exam (X) \(\to\)
Result (Y) \(\downarrow\)		 PSC JSC SSC HSC Total
Passed 30 26 23 25 104
Failed 12 13 10 14 49
Absent 5 2 3 4 14
Total 47 41 36 43 167
  • Marginal: \(P(Pass) = P(x_1)=P(x_1,y_1)+P(x_1,y_2)+P(x_1,y_3)\)
  • \(P(Absent) = P(x_3) = P(x_3,y_1)+P(x_3,y_2)+P(x_3,y_3)\)

Marginal Probability

Consider the previous table

Joint probability: \(P(x_i, y_j); i = 1,2, \cdots m; j = 1,2, \cdots n\)

Marginal probability \(\to P(x_i) \leftarrow P(x_i, y_j)\)

  • For x: \(P(x_i) = \sum_{j=1}^n P(x_i, y_j); i = 1,2, \cdots m\)
  • For y: \(P(y_i) = \sum_{i=1}^m P(x_i, y_j); j = 1,2, \cdots n\)
  • What about continuous x?

Marginal Probability Properties

  • \(P(x_i) \ge 0\) and \(P(y_i) \ge 0\)
  • \[\sum_{i=1}^m P(x_i)=\sum_{j=1}^n P(y_j)=1\]

Summing marginal probabilities will give 1.

Joint PMF Example

\(P(x,y) = \frac{x+y}{9}; x=0,1,2; y = 0, 1\)

  • Find marginal probabilities
  • Check properties (sum)
  • \(P(x) = \frac{2x+1}{9}\)

Joint PDF Example

\(f(x,y) = \frac{2x+y}{3}; 0 \le x \le 1.5\) and \(0 \le y \le 1\)

Conditional Probability Function

Like Bayes Theorem

\(P(X_i|y_j) = \frac{P(x_i,y_j)}{P(y_j)}; P(y_j) \gt 0\)

Properties

  • \(\sum_{j=1}^m P(x_i|y_j)=\sum_{i=1}^m P(y_j|x_i)=1\)

Conditional Probability Example

\(P(x,y) = \frac{x+y}{9}; x=0,1,2; y = 0, 1\)

Find \(P(X|Y)\) and \(P(Y|X)\)

Find for continuous X as well.

Find k for pdf

\(f(x) = kx^2+kx+\frac 1 8; 0 \lt x \lt 2\)

  1. Find k
  2. Find \(P(1 \lt X \lt 2)\)

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