*(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] *= a^{2} 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

σ^{2} = 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 t^{r}/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*

µ_{3} = µ‘_{3} - 3 µ‘_{2}. µ + 2 µ^{3}

*, x<0*

*Find the distribution function, mean and variance.*

** **

(e) **Skewness and Kurtosis.**

Skewness =

For symmetric distribution, β_{1} = 0. If β_{1} > 0 then the distribution is called positively skewed.

If β_{1} < 0, then the distribution is called negatively skewed.

For , β_{2 }= 3 the distribution is called platykurtic.

β_{2 }> 3 the distribution is called mesokurtic.

β_{2 }< 3 the distribution is called leptokurtic.

** **

**Example 4. ***A continuous random variable X has a pdf
*

*f(x) *= *3x ^{2}, 0 *<

*x*<

*1*

*Obtain the first four central moments and hence calculate *β_{1}* and *β_{2}

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 *= *x ^{1} *=

*q*=

^{2}p, 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.