Random VariablesMr. Rohan Bhatkar

Content-

Random VariablesProbability Mass FunctionDiscrete Random VariablesDistribution FunctionVarianceExpectationContinuous Random Variables

Random Variables

Consider an experiment of throwing two dice. We know that this experiments has 36 outcomes .Let x be the sum of numbers on the uppermost faces.

For Example-The value of x is 2 if outcome of the experiment is (1,1) & probability of this outcome is 1/36.

In this case we can associate real number with each outcome of random experiment or group of outcomes of random experiment.Here X is called Random Variables.

Probability Mass Function

If x is discrete random variables values x1, x2, x3, …., xn then probability of each value is described by a function called the Probability Mass Function. The probability that random variable X takes values xi is denoted by p(xi).

Illustration-Suppose fair coin is marked 1 & 2, dice numbered 1, 2, 3, 4, 5 & 6 are thrown simultaneously then probability mass function of random variables X which is sum of numbers on coin & dice is obtained as under.The sample space is:

S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}Note that n(s) = 12X: sum of numbers on coin & dice.

X = 2, 3, 4, 5, 6, 7, 8P[x=2] = {(1,1)} = 1/12P[x=3] = {(1,2)(2,1)} = 2/12P[x=4] = {(1,3)(2,2)} = 2/12P[x=5] = {(1,4)(2,3)} = 2/12P[x=6] = {(1,5)(2,4)} = 2/12P[x=7] = {(1,6)(2,5)} = 2/12P[x=8] = {(2,6)} = 1/12

Thus probability distribution of random variable X is as under:X=x 2 3 4 5 6 7 8

P[X=x] 1/12 2/12 2/12 2/12 2/12 2/12 1/12

Note that P[xi] ≥ 0 & ∑ (xi) = 1

Discrete Random Variable

A random variable which can assume only a countable number of real values & the values taken by variables depends on outcome of random experiment is called Discrete Random Variables.

For Example-1. Number of misprints per page of book.2. Number of heads in n tosses of fair coin.3. Number of throws of dice to get first 6.

Distribution Function

Let X be a random variable. The function F defined for all real values x by F(x) = P [X = x], For all real x is called Distribution Function.A distribution function is also called as Cumulative Probability Distribution Function.

Properties OF Distribution Function:

If F is the D.F. of random variable X & if a <b thenP[a < X ≤ b] = F(b) – F(a).

Values of all distribution functions lie between 0 & 1i.e. 0 ≤ F(x) ≤ 1 for all x.

All distribution functions are monotonically non-decreasingi.e. 0 < y then F(x) < F(y).

F(- ∞) = lim F(x) = 0x - ∞

F(+ ∞) = lim F(x) = 1x → ∞

If X is discrete random variable then F(x) = ∑ P(xi) xi ≤ x

If values of discrete random variable X are like x1 < x2 < x3 < x4 … then P(Xn+1) = F(Xn+1) – F(Xn)

If x is discrete random variable then D.E. is step function.

Illustration-Consider probability distribution of random variable x

If F(x) is distribution function of random variable x thenF(1) = P[x ≤ 1] = P(1) = 0.1F(2) = P[x ≤ 1] = P(1) + P(2) = 0.1 + 0.2 = 0.3F(3) = 0.1 + 0.2 + 0.3 = 0.6F(4) = 0.1 + 0.2 + 0.3 + 0.2 = 0.8F(5) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1 = 0.9F(6) = 1

Thus values of x & corresponding cumulative probability distribution function is as under:

X=x 1 2 3 4 5 6

P[X=x] 0.1 0.2 0.3 0.2 0.1 0.1

X=x 1 2 3 4 5 6

P[X=x] 0.1 0.3 0.6 0.8 0.9 1.0

We will have graphical representation of random variable X, & the Graph of D.E.

1 2 3 4 5 6

1

2

3

X →

P(x)

Probability Distribution

1 2 3 4 5 6

0.2

0.4

0.6f(x)

CumulativeProbability Distribution

0.8

1.0

X →

Variance

If X is discrete random variable then variance of x is given byVar (X) = E[X – X (X)]2

Note that Var (X) = µ2Var (X) = µ2

` - µ1`2

Var (X) = E (X2) – [E(X)] 2 The positive square root of the variance is known as standard deviation

S D (X) = + √Var (X)

Illustration-For a random variable X, E (X) = 10 and Var (X) = 5. Find Var (3X + 5), Var(X – 2), Var (4X).

Also find E(5X – 4), E(4X + 3).

Solution-Var (3X + 5) = Var (3X)= 9 Var (X) = 9 x 5 = 45Var (x - 2) = Var (X) = 5Var (4X) = 16 Var (X) = 16 x 5 = 80Var (5X - 4) = 5 E(X) – 4Var (5X - 4) = 5 x 10 – 4 = 46Var (4X + 3) = 4 E(X) + 3Var (4X + 3) = 4 x 10 + 3 = 43

Properties OF Variance:

1. Variance is independent of change of origin. It means that if X is random variable & ‘a’ is constant then variances of X and new variable X + a are same i.e.

Var (X) + Var (X + a)consider

Var (X + a) = E [X + a – (E (X) + a)]2 Var (X + a) = E [X + a – E (X) + a]2 Var (X + a) = E [X –E (X)]2 Var (X + a) = Var (X).

2. Variance of random variable depends upon change of scale.i.e. Var (a X) = a2 Var (X)consider

Var (a X) = E [aX – E (aX)]2

= E [aX – a E (X)]2 = E a2 [X –E (X)]2

= a2 E [X –E (X)]2

= a2 Var (X).3. From property 1 and 2 we can find Var (a X + b),

Var (a X + b) = Var (a X) by property 1 = a2 Var (X) by property 2

4. Variance of constant is 0. put a=0 in a x + b thenVar (a X + b)= Var (b)

but Var (a x + b) = a2 × Var (X)Var (a x + b) = 0 × Var (X) = 0Var (b) = 0

ExpectationMathematical Expectation of Discrete Random Variable:

Once we have determined probability distribution function P(x) & distribution function of discrete random variable X, we want to compute the mean or variance of random variable X, the mean or expected value of X is nothing but weighted average of value X where corresponding probabilities are taken as weights, thus if X takes values x1, x2,.. With corresponding probabilities

p(x1), p(x2),… then mathematical expectation of X denoted by E(X) is given by E(X) = ∑ xi P(xi).E(X) is also mean of random variable X.E(X) exists if series on right hand side is absolutely convergent.

Illustration-The p.m.f. of a random variable X is given below. Find E(X).Hence E(2X + 5) & E(X - 5).

X=x 1 2 3 4 5 6

P[X=x] 0.1 0.15 0.2 0.3 0.15 0.1

Solution-Expected value of random variable is given by:

E(X) = ∑xi P[X = xi] = 1(0.1) + 2(0.15) + 3(0.2) + 4(0.3) + 5(0.15) + 6(0.1) = 0.1 + 0.3 + 0.6 + 1.2 + 0.75 + 0.6E(X) = 3.55

E(2X + 5) = 2 E(X) + 5 = 2 E(3.55) + 5

= 12.1 E(X - 5) = E(X) - 5 = (3.55) - 5

= -1.45

Properties OF Expectation:

If X1 & X2 are two random variables then E(X1 + X2) = E(X1 ) + E(X2). This result can be generalized for X1, X2, …, Xn i.e. n random variables

E(X1 + X2 + …. + Xn ) = E(X1) + E(X2) + …. + E(Xn)

If X & Y are independent random variable, E(XY) = E(X) E(Y) If X is a random variable & a is constant then

E(a X) = a E(X) andE(X + a) = E(X) + a

If X is a random variable and a & b are constantsE[a (X) + b] = a E(X) + b

If X is a random variable and a & b are constants & g (X) a function X is random variable thenE[a g (X) + b] = a E[g (X)] + b

If X1, X2, X3 ….,Xn are any n random variable and if a1, a2, ….,an are any n constants thena1 X1 + a2 X2 + …. An Xn is called linear combination of n variables & expectation of linear combination is given by E(a1 X1 + a2 X2 + ….. + an Xn)

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If X ≥ 0 then E(X) ≥ 0. If X and Y are two random variables & if X ≤ Y then E(X) ≥ E(Y). If X & Y are independent random variable then

E[a g (X) + h (Y)] = E[g (X)] E[h (Y)]then g (X) is a function of X & its random variable. Also h (Y) is a function of Y & random variable.

Continuous Random Variables

Consider the small interval (X, X + dx) of length dx round the point x.Let f(x) be any continuous function of x so that f(x) represent the probability that falls in very small interval (X, x + dx)Symbolically P[x ≤ X ≤ x + dx] = f(x) dx.

X – dx/2

f(x) dx

X + dx/2

Y = f(x)

x

y

In the figure f(x) dx represents the area bounded by the curve y = f(x); X-axis a& the ordinates x and x + dx.The function f(x) so defined is known as probability density function of random variable X & usually abbreviated as p.d.f. The expression f(x) dx usually written as F (x), is known as probability differential & f(x) is known as probability density curve. The probability that X lie in the interval dx is f(x) dx. Thus the p.d.f. of random variable x is defined as

f(x) = lim P [x ≤ X ≤ x + δx] δx → 0 δx

Illustration-A random variable X has following p.d.f.

f(x) = K - ∞ < x < ∞ 1 + x2 = 0 OtherwiseFind k:

Solution:Since X is continuous random variable with density function f(x),

∞ ⌠ f(x) dx = 1 - ∞

∞ ⌠ K dx = 1 - ∞ 1 + x2

∞k[tan-1 x] = 1 - ∞

k[tan-1 ∞ - tan-1 ∞ (- ∞)] = 1

k[π/2 + (π/2)] = 1

k × π = 1

k = 1/ π

Thank You …