Question

numbers of labour and capital equal to each other? 5. (5 + 5 = 10 points) (1) Does a continuous random variable X exist with E(X) = 2? If yes, give an example for its probability density function. If no, give an explanation. (2) Do random variables Y and Z exist with Y = aZ for a constant a, p(Y,Z) = -1, E(Y) = 9, and E(Z) = 3? If yes, compute the value of a. If no, give an explanation.

Answer 1

Solution: If x is a continuous random variable Whose cumulative distribution function admits a density f(a), ie. the pdf is f(x); then the expected value is defined as the Lebesque integrale - E(x) = (x f(x) dx R Now there can be unmimbered instances where the a Rov. X has an expectation, E(X)=2. So the answer is res, and in Qeveral of examples of 6 are there whose which follow the properties of being a pdf and also gives reault=2 for the integral stated above on 4 on its range. Example: let us define a Rovex with its par f(x) as, f(x)= 3 if o {x24 Sodo rhoda o otherwise E(x)= (x fagax theat E(x) = 2]

- We know, if x is a Random Variable a in some constant then le= Eax) = a E(X)=(so 3 and 8 = 1 Now, given are two Random Variables Y and Z with Y= az for a constant a. role were Now P (Y,Z) = COV (Y,Z) Two 2 ✓ Var (Y) Var (7) - Covca Z, I) ✓ Var (az) Var (2) Griven - to PCY, Z) = -1 - cov Caz, z) Var(az) Vaz (2) - a cor (2, 2) v az var (2) Var (2) as a vart2) - -> .-1. = a has a negative value Y=aZ given, lai varge) = -1 o NOW

Continued from previous page, E(Y) = 9 } briven in de word she E(Z) =3 J 9=E(Y) = E(az)= a E(z) = (x3 = 3a. i a=3 - fü) Now is and C are contradictory Conclusions. So our conclusion is No such Randem variables. Y and I do not exist.