Properties of Mathematical expectation and variance

**Properties
of Mathematical expectation and variance**

(i) *E*(*aX*
+ *b*)* = aE*(* X *)*
+ b *,* *where* a *and* b *are constants

**Proof**

Let *X* be a
discrete random variable

Similarly, when *X*
is a continuous random variable, we can prove it, by replacing summation by integration.

(ii) *Var
*(* X *)* *=* E *(* X*^{2}* *)*
*−(* E *(* X *))^{2}

We know

*E *(* x*) = *μ*

*Var *(*
X *) = E (*X* – *μ*)^{2}

= E (*X ^{2}*
– 2

= E (*X ^{2}*)
– 2

(Since μ is a constant)

= E (*X ^{2}*)
– 2

*Var *(*
X *) = E (*X ^{2}*) – (E (

An alternative formula to compute variance of a random
variable* X *is

* σ*^{2}* *=* *Var* *(* **X** *)* = **E*(* **X*^{2})* *–
(*E*(* **X** *))^{2}

(iii) *Var(aX
+b) = a ^{2}Var *(

*Var *(*
aX *+* b*)* *=* E (*(*
aX* + *b*)* *– *E *(* aX + b*))^{2}

*= E ( aX + b *– *aE*(* X *)* *– *b*))^{2}

*= E *(*aX
*– *aE *(* X *))* *^{2}

*= E *(* a *(* X *−*E*(* X *)))^{2}

= *a*^{2} *E *(*X* – *E*(*
X *))^{2}* *.

Hence *Var *(* aX *+* b*)*
*=* a*^{2}*Var *(*
X *)

Variance
gives information about the deviation of the values of the random variable
about the mean *μ*. A smaller *σ* ^{2} implies that the
random values are more clustered about the mean, similarly, a bigger *σ* ^{2} implies that the
random values are more scattered from the mean.

The above figure shows the pdfs of two continuous random
variables whose curves are bell-shaped with same mean but different variances.

** **

**Example 11.16**

Suppose that *f* ( *x*) given below represents a probability
mass function,

Find (i) the value of c (ii) Mean and variance.

**Solution**

(i) Since *f *(* x*)*
*is a probability mass function,* f *(*x*)*
*≥* *0 for all *x *, and ∑_{x}*f *(*x*)
= 1* *.

Since *f *(*x*)*
*≥* *0* *for all *x *, the possible value of* c *is 1/5.

Hence, the probability mass function is

(ii) To find mean and variance, let us use the following
table

Therefore the mean and variance are 4.6 and 2.24
respectively.

** **

Two balls are chosen randomly from an urn containing 8 white
and 4 black balls. Suppose that we win Rs 20 for each black ball selected and
we lose Rs10 for each white ball selected. Find the expected winning amount and
variance.

Let *X* denote the
winning amount. The possible events of selection are (i) both balls are black, or
(ii) one white and one black or (iii) both are white. Therefore* x *is a random variable that can be defined
as

* X*
(both are black balls) = ₹ 2(20) =
₹
40

* X *(one
black and one white ball)* *=* *₹* *20* *−* *₹* *10* *=* *₹10

*X* (both are white balls) = ₹ (− 20) = − ₹ 20

Therefore *X* takes
on the values 40,10 and −20

Total number of balls *n
= 12*

** **

**Example 11.18**

Find the mean and variance of a random variable *X* , whose probability density function
is

Tags : Probability Distributions | Mathematics , 12th Maths : UNIT 11 : Probability Distributions

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12th Maths : UNIT 11 : Probability Distributions : Properties of Mathematical expectation and variance | Probability Distributions | Mathematics

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