# If two random variables have the joint probability density function$$f(x, y) = \left\{\begin{matrix}Ke^{-(x+y)}, \enspace 0 < x < ∞, \enspace 0 < y < ∞\\ 0, otherwise\end{matrix}\right.$$Find (i) constant k (ii) Conditional probability density function of X and given Y (iii) Var(3X + 2Y)

Solution:

Given,

Joint probablity density function

f(x, y) = $$\left\{\begin{matrix}ke^{-(x+y)}, 0 < x < ∞, o < y < ∞\\ 0, otherwise\end{matrix}\right.$$

Solution i):

We know,

$$\int_0^∞ \int_0^∞ ke^{-(x+y)} dx dy = 1$$

$$\int_0^∞ \int_0^∞ ke^{-x} . e^{-y} dx dy = 1$$

$$\int_0^∞ k . e^{-y} \left \{ \int_0^∞ e^{-x} dx \right \} dy = 1$$

$$\int_0^∞ k . e^{-y} \left [ -e^{-x} \right ]_0^{∞} dy = 1$$

$$\int_0^∞ k . e^{-y} \left [ -\frac{1}{e^∞} + 1 \right ] dy = 1$$

$$\int_0^∞ k . e^{-y} dy = 1$$

$$k . [-e^{-y}]_0^∞ = 1$$

k = 1

Solution ii):

we know,

$$f\left ( \frac{x}{y} \right ) = \frac{f(x, y)}{f(y)}$$

$$= \int_0^∞ e^{-(x+y)} dx$$

$$= \int_0^∞ e^{-x} . e^{-y} dx$$

$$= e^{-y} \int_0^∞ e^{-x} dx$$

$$= e^{-y} \left [ -e^{-x} \right ]_0^∞ dx$$

= e-y

∴ $$f\left ( \frac{x}{y} \right ) = \frac{e^{-(x+y)}}{r^{-y}}$$

= $$\frac{e^{-x} . e^{-y}}{e^{-x}}$$

= e-x

Solution iii):

$$f(x) = \int_0^∞ e^{-(x+y)} dx$$

= $$e^{-x} \int_0^∞ e^{-y} dy$$

= $$e^{-x} \left [ -e^{-y} \right ]_0^∞$$

= e-y

Now,

E(x) = $$\int_0^∞ x f(x) dx$$

= $$\int_0^∞ x e^{-x} dx$$

= 1

E(x2) = $$\int_0^∞ x^2 f(x) dx$$

= $$\int_0^∞ x^2 e^{-x} dx$$

= 2

Also,

E(y) = 1

E(y2) = 2

We know,

var(3x + 2y) = $$3^2 \times var(x) + 2^2 \times var(y)$$

= $$9\left [ E(x^2) – \{E(x)\}^2 \right ] + 4 \times \left [ E(y^2) – \{E(y)\}^2 \right ]$$

= 9[2 – 1] + 4[2 – 1]

= 9 + 4

= 13