91Ó°ÊÓ

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

(a):

There are two ways to compute the probability. First is to compute marginal pdf of \({\rm{X}}\) and then find \({\rm{X > 3}}\) using the marginal pdf, or compute it immediately. Since we need to determine marginal pdf's of \({\rm{X}}\) and \({\rm{Y}}\) in part\({\rm{(b)}}\), we will show you the second way (in which you indirectly compute pdf of \({\rm{X}}\) ).

For every adequate set \({\rm{A}}\) the following holds

The following holds

\(\begin{aligned}{l}{\rm{P(X > 3) = }}\int_{\rm{3}}^{\rm{Â¥}} {\overbrace {\int_{\rm{0}}^{\rm{Â¥}} {\rm{x}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dy}}}^{{\rm{marginal pdf of X}}}} {\rm{dx}}\\{\rm{ = }}\int_{\rm{3}}^{\rm{Â¥}} {\int_{\rm{0}}^{\rm{Â¥}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x}}}}{\rm{ \times }}{{\rm{e}}^{{\rm{ - xy}}}}{\rm{dydx}}\\{\rm{ = }}\left. {\int_{\rm{3}}^{\rm{Â¥}} {\rm{x}} {{\rm{e}}^{{\rm{ - x}}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{x}}}} \right){{\rm{e}}^{{\rm{ - xy}}}}} \right|_{\rm{0}}^{\rm{Â¥}}\\{\rm{ = }}\int_{\rm{3}}^{\rm{Â¥}} {{{\rm{e}}^{{\rm{ - x}}}}} {\rm{dx = - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{3}}^{\rm{Â¥}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 3}}}}{\rm{ = 0}}{\rm{.05}}\end{aligned}\)

(b):

The marginal probability density function

of continuous random variable \({\rm{X}}\) is

\({{\rm{f}}_{\rm{X}}}{\rm{(x) = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} {\rm{(x,y)dy, for - ¥< x < ¥}}\)

The marginal probability density function

of continuous random variable \({\rm{Y}}\) is

\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} {\rm{(x,y)dx, for - ¥< y < ¥}}{\rm{.}}\)

The marginal pdf of \({\rm{X}}\) is

We can write it as

Similarly, the marginal pdf of \({\rm{Y}}\) is

\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\int_{{\rm{ - ¥}}}^{\rm{¥}} {\rm{f}} {\rm{(x,y)dx = }}\int_{\rm{0}}^{\rm{¥}} {\rm{x}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx}}\)

\({\rm{ = }}\left| {\begin{aligned}{*{20}{c}}{{\rm{u = x}}}&{\rm{n}}&{{\rm{du = dx}}}\\{{{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx = dv}}}&{\rm{n}}&{{\rm{v = - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}}\\{{\rm{ integration by parts: }}}&{{\rm{uv}}}&{{\rm{ - }}\int {\rm{v}} {\rm{du}}}\end{aligned}} \right|\)

\({\rm{ = }}\left. {{\rm{x \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} \right)} \right|_{\rm{0}}^{\rm{Â¥}}{\rm{ + }}\int_{\rm{0}}^{\rm{Â¥}} {\frac{{\rm{1}}}{{{\rm{1 + y}}}}} {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dx}}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{0 + }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}\int_{\rm{0}}^{\rm{Â¥}} {{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} {\rm{dx}}\)

\({\rm{ = }}\left. {\frac{{\rm{1}}}{{{\rm{1 + y}}}}{\rm{ \times }}\left( {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{1 + y}}}}{{\rm{e}}^{{\rm{ - x(1 + y)}}}}} \right)} \right|_{\rm{0}}^{\rm{Â¥}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{{{\rm{(1 + y)}}}^{\rm{2}}}}}\),\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = 0}}\),,\({\rm{y < 0}}\).

(1): here we used L'Hopital's rule to obtain the limit:

where \({\rm{c}}\) is constant we get when we derivative the denominator.

The marginal pdf of \({\rm{Y}}\) is

Two random variables \({\rm{X}}\) and \({\rm{Y}}\) are independent if and only if

1. \({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) discrete rv's,

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\),

for every \({\rm{(x,y)}}\) and when \({\rm{X}}\) and \({\rm{Y}}\) continuous rv's,

otherwise, they are dependent.

It is obvious that the joint pdf is not the product of marginal pdf's for every\({\rm{(x,y)}}\), hence, random variables are dependent.

(c):

At least one component exceeds \({\rm{3}}\) means that we need to find probability of event

\({\rm{\{ X > 3\} `E \{ Y > 3\} }}\)

However, it is easier if we look at the complement of the mentioned event:

\({\rm{\{ X£3\} {C}\{ Y£3\} }}\)

Therefore, the following holds

\(\begin{aligned}{l}{\rm{P(\{ X > 3\} `E \{ Y > 3\} ) = 1 - P(\{ X£3\} {C}\{ Y£3\} )}}\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{1 - }}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{\rm{3}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x(1 + y)}}}}{\rm{dydx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {\int_{\rm{0}}^{\rm{3}} {\rm{x}} } {{\rm{e}}^{{\rm{ - x}}}}{{\rm{e}}^{{\rm{ - xy}}}}{\rm{dydx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {\rm{x}} {{\rm{e}}^{{\rm{ - x}}}}\left( {{\rm{ - }}\frac{{\rm{1}}}{{\rm{x}}}\left( {{{\rm{e}}^{{\rm{ - 3x}}}}{\rm{ - }}{{\rm{e}}^{{\rm{ - x \times 0}}}}} \right)} \right){\rm{dx}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{\rm{3}} {{{\rm{e}}^{{\rm{ - x}}}}} \left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - 3x}}}}} \right){\rm{dx}}\\{\rm{ = 1 - }}\left( {{\rm{ - }}\left. {{{\rm{e}}^{{\rm{ - x}}}}} \right|_{\rm{0}}^{\rm{3}}} \right){\rm{ + }}\left( {{\rm{ - }}\left. {\frac{{\rm{1}}}{{\rm{4}}}{{\rm{e}}^{{\rm{ - 4x}}}}} \right|_{\rm{0}}^{\rm{3}}} \right)\\{\rm{ = 0}}{\rm{.3}}\end{aligned}\)

(1): for every adequate set \({\rm{A}}\) the following holds