Let (X, Y) be jointly distributed with density f(x, y) = e^(-y), for 0 < x < y < ∞ ISS PREVIOUS YEAR PAPER-1 2016

ISS PREVIOUS YEAR PAPER-1 2016 SET A Q.NO.-1

Question

Let (X, Y) be jointly distributed with density

f(x, y) = e^(-y), for 0 < x < y < ∞ OR f(x, y) = 0, otherwise

Consider the following statements:

1. E(X) = 1

2. E(Y) = 2

3. E(XY) = 2

Which of the above are correct?

(a) 1 and 2 only(b) 2 and 3 only(c) 1 and 3 only(d) 1, 2 and 3

Solution

The joint probability density function is

f(x, y) = e^(-y), where 0 < x < y < ∞

1. Calculation of E(X)

E(X) = ∫ from 0 to ∞ ∫ from 0 to y x e^(-y) dx dy

Since e^(-y) does not depend on x,

E(X) = ∫ from 0 to ∞ e^(-y) [ ∫ from 0 to y x dx ] dy

Now,

∫ from 0 to y x dx = y² / 2

Therefore,

E(X) = ∫ from 0 to ∞ e^(-y) (y² / 2) dy

E(X) = (1/2) ∫ from 0 to ∞ y² e^(-y) dy

Using the Gamma integral,

∫ from 0 to ∞ y² e^(-y) dy = 2

Therefore,

E(X) = (1/2) × 2 = 1

So, statement 1 is correct.

2. Calculation of E(Y)

E(Y) = ∫ from 0 to ∞ ∫ from 0 to y y e^(-y) dx dy

E(Y) = ∫ from 0 to ∞ y e^(-y) [ ∫ from 0 to y dx ] dy

Now,

∫ from 0 to y dx = y

Therefore,

E(Y) = ∫ from 0 to ∞ y² e^(-y) dy

We know,

∫ from 0 to ∞ y² e^(-y) dy = 2

Thus,

E(Y) = 2

So, statement 2 is correct.

3. Calculation of E(XY)

E(XY) = ∫ from 0 to ∞ ∫ from 0 to y xy e^(-y) dx dy

E(XY) = ∫ from 0 to ∞ y e^(-y) [ ∫ from 0 to y x dx ] dy

We know,

∫ from 0 to y x dx = y² / 2

Therefore,

E(XY) = ∫ from 0 to ∞ y e^(-y) (y² / 2) dy

E(XY) = (1/2) ∫ from 0 to ∞ y³ e^(-y) dy

Now,

∫ from 0 to ∞ y³ e^(-y) dy = 3! = 6

Thus,

E(XY) = (1/2) × 6 = 3

So, statement 3 is incorrect.

Final Answer

Only statements 1 and 2 are correct.

Correct option: (a) 1 and 2 only

ISS PREVIOUS YEAR PAPER-1 2016 SET A Q.NO.-1

Click Here to download pdf