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Probability density function (pdf): Let ‘ X ’ be a continuous random variable taking values in the interval [a, b], then, A function f(x) is said to be the probability density function of the continuous random variable ‘ X ’, if it satisfies the following conditions :

•  f(x) ≥ 0 for all ‘ X’ in the interval [a, b] .
• For two distinct numbers c & d in the interval [a, b]:

P(c ≤ X ≤ d) = (Area under the probability curve between ordinates at X =c and X=d).

• Total area under the curve is 1. i.e. , P(- ∞ < x < ∞) = 1 .

• E(X) = XΣX.P (X) 
• E(a) = a, 
• E(aX) = a E(X) and 
• E(aX+b) = aE(X)+b.

• Var(X)= {E(X² )-[E(X)]²},
• V (a)=0,
• V(aX)=a² V(X)
• V(aX+b)=a²V(X), and S.D.(X)= √v(x).

We know S.D(x)

V(2x) = 2²V(x) 

= 4 (4) = 16; 

Since V(aX) 

= a² V(X).

We know that E(X+Y) = E(X) + E(Y) ; 

7 = 4 + E(y) 

E(y) = 7 – 4 = 3.

Given E(-x/4) 

= E(-1/4. x) 

=-1/4.E(X) 

= -1/4. 2 

=-1/2 since E(aX) 

= a E(X)

V(-5X) 

= (5)²V(x) 

= 25(5) 

= 125; 

Since V(aX)

=a²V(X).

V(5x-9)this is of the type V(aX+b)=a²V(X)

V(5x-9) = 5². V(x)

Here V(X) = E(X²) – [E(x)]² ;

= 25 – 42 = 25 – 16 = 9

∴ V(5×-9) = 5² . V(x) = 25(9) = 225

V(2X-5) = 2² V(x) 

= 4 (1.6) 

= 6.4; 

Since V(aX+b)=a² V(X),

E(3X + 2Y) = 3 E(x) + 2 E(y) = 3(3) + 2(5) = 9 + 10=19 

E(3X – Y) = 3 E(X) + (-1) E(Y) = 3(3)- 5 = 9 – 5 = 4

Given E(5x/3) 

= E(5/3.x) 

= 5/3.E(x) 

= 5/3.2/5 

= 2/3

E(3X – 6) 

= 3E(x) – 6 

= 3(2) – 6 

= 6 – 6 

= 0

E(3 + 4X) 

= E(4x + 3) 

= 4.E(x) + 3 

= 4.(1.5) + 3 

= 6 +3 = 9 

since E(ax+b) 

= aE(x)+b.

We know for any two independent random variables: 

E(XY) = E(X)E(Y);

10 = 4 E(Y); E(Y)= 10/4 = 2.5

We know that Cov(X, Y) = E(XY) – E(X).E(Y)

= 11 – (5) (2)=11 -10= 1.

A random variable which assumes all the possible values in its range is called a continuous random variable.

A systematic presentation of the values taken by a random variable with respective probabilities is called the probability distribution of a random variable’.

Let ‘X’ be a discrete random variable which can takes the values x₁, x₂, x₃,… ,xn with respective probabilities p₁, p₂,p₃,… , pn then, the mathematical expectation of ‘ X ’ is:E(X) = x₁p₁ + x₂ P₂ + x₃ P₃ + – + xnPn

Then E(X) = Σ x p(x) ;it is also called as mean of discrete random variable (X)

Statement: Let X and Y be two independent random variables with respective expectations E(X) and E(Y). Then expectation of the product of these random variables is; E(XY) = E(X) E(Y).

Cov(X, Y) =E(XY) – E(X).E(Y),

Random variable is a function which assigns a real number to every sample point in the sample space.

Proof: From the definitions of probability mass function and mathematical expectation

1. E(a) = Σa .P(a) = a. ΣP(x) = a.1 = a. 

∵ ΣP(x) = 1

2. E(ax) = Σax P(x) = aΣx P(x) = a. E(x) 

∵Σx P(x) = E(x)

3. E(ax + b) = E(ax + b). P(x) Removing bracket

= Σax. P(x) + Σb.P(x) = aΣx . P(x) + bΣP(x)

= a. E(x) + b. 1 = aE(x) + b

4. Var (a) = E[a – E(a)]² ; 

∴ V(x) = E[x – E(x)]²

= E[a – a]² = E(0) = 0

5. Var (an) = E[ax – E(ax)]² = E[ax – aE(x)]² ; 

∴E(ax) = aE(x)

= a²E(x – E(x)]² = a² Var (x).

We know that V(X) = E(X² )-[E(x)]²

14 = 50-[E(x)]²

[E(x)]² = 50 – 14 = 36

Squaring on both sides, we get E(x) =√36 = 6

A random variable ‘ X ’ which takes the specified values x₁ x₂,…,x_n with a respective probabilities p₁, p₂,….., p₁ is a discrete random variable.

V(3X+4) = 3² V(x) 

= 9 (4) 

= 36;

V(-3X) = (-3)² V(x) 

= 9 (4) 

= 36; 

Since V(aX+b)

=a² V(X),

Statement: Let X and Y be two random variables with respective expectations E(X) and E(Y). Then, E(X + Y)=E(X) + E(Y).