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Characteristics of Probability Distributions

(a) Distribution function. For discrete case, the distribution function denoted by F(x) is defined as

 

                             F(x) = P[X ≤ x]

 

          For the above example, the distribution function is given by

For continuous case, the distribution function is defined as

For the above example,

Properties:

(i) Discrete case

P (a < X < b) =  F(b) - F(a)

P (a ~ X < b) = P[X = a] + [F(b) - F(a)]

P (a < X ~ b) = [F (b) - F (a)] - P[X = b] 

P (a < X < b) = F (b) - F (a) + P [X = a] - P [X = b]

 

Continuous case

P (a < X < b) = P (a < X < b) = P (a < X ~ b) = P (a < X < b) = F(b) - F(a).

(ii)

(iii) F(x) ->F(y) whenever x < y.

(iv) F(a) - F(a - 0) = P[x = a] and F(a + 0) = F(a)

(v) For continuous case, F'(x) = j{x) ~ 0 => F(x) is non-decreasing function.

 

(b) Mean/expectation. Let X be the random variable. Then mean/expectation is defined as

(Discrete case)                              (Continuous case)

This expectation is sometimes called as 'Population Mean'.

 

Properties:

(i) E[X + Y] = E[X] + E[Y]

(ii) E [cX] = c E[X]

(iii) E[c] = c and E [X + c] = E[X] + c

(iv) If X and Y are independent then E (XY) = E(X). EM

(v) Physically, expectation represents the center of mass of the probability distribution.

 

(c) Variance. Variance of a probability distribution is given by

 

Properties :

(i) V[aX + b] = a2 V[X]

(ii) Physically variance represents the moment of inertia of the probability mass distribution about a line through the mean perpendicular to the line of the distribution.

Example 1. For the following distribution

 

(i)Find the value of k.

(ii)Find the mean and variance.

(iii)Find the distribution function.

 

            Solution. (i) Since this is a pmj we have

Σ p(x) =1

=> 0.1 + k + 0.2 + 3k + 0.3 = 1

=> 4k + 0.6 = 1

=> 4k = 0.4

=> k = 0.1.

 

(ii) µ = Mean = Σ x. p(x)

= 1(0.1) + 2(0.1) + 3(0.2) + 4(0.3) + 5(0.3) = 3.6

σ2 = Varience = Σ (x - µ)2 . p(x)

=(1 - 3.6? (0.1) + (2 - 3.6)2 (0.1) + (3 - 3.6)2 (0.2)+ (4 - 3.6f (0.3) + (5 - 3.6)2 (0.3)

=1.64.

 

(iii) Distribution function is given as follows :

Example 2. Find the mean, variance and distribution function of the pdf

            f(x) = ax2, 0 .::;, x .::;, 1

= 0,      elsewhere.

 

·Solution. Since this is a pdf, then

a = 3

 

So the pdf can be taken as

 

            f(x) 3x2,           0 < x > 1

= 0, elsewhere

Distribution function

(d) Moments and moment generating function.  rth moment about the mean is defined as

Now consider the discrete case,

denoting p(x) by P

Therefore,

µ’ = Coefficient of tr/r! in the expansion of Mx(t).

Alternatively, we have

t=0

The relations between the central moments and raw moments are as follows:

 

The relations between the central moments and raw moments are as follows:

µ2 = µ’2 - µ2 = Variance

(The moments obtained from a distribution (discrete/continuous) is called "Population  oments'').

 

Note. I. For m.gf if the point a is not given, it is taken as zero.

 

2. A r.v. X may have no moments although its m.g.f exists.

e.g                     f(x) = 1/(x+1)(x+2),   x=0,1,2…… (The reader can verify)

 

3. A r.v. X may have moments although its m.gf fail to generate the moments.

Example 3. The pdf of Rayleigh distribution is given by

µ3 = µ‘3 - 3 µ‘2. µ  + 2 µ3

, x<0

Find the distribution function, mean and variance.

 

(e) Skewness and Kurtosis.

 

Skewness   =  

For symmetric distribution, β1 =  0. If  β1  > 0 then the distribution is called positively skewed.

If  β1  < 0, then the distribution is called negatively skewed.

 

For , β2 = 3 the distribution is called platykurtic.

β2 > 3 the distribution is called mesokurtic.

β2 < 3 the distribution is called leptokurtic.

 

Example 4. A continuous random variable X has a pdf

f(x) = 3x2, 0 < x< 1

Obtain the first four central moments and hence calculate β1 and β2

β2 = µ4 / µ22 = 0.00435/(0.0375)2 = 3.09

Since β2 > 3, the distribution is leptokurtic.

 

Example 5. Find the moment generating function of the following distribution.

P[X = x1 = q2p, x = 0,1,2, ..... , 0 < p .:S 1, q = .1 - p

= 0, otherwise.

Hence find the mean and variance.

Solution. The given distribution is a discrete distribution.