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)

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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

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