Determine Whether the Function Is a Probability Density Function Over$$f ( x ) = \left\{ \begin{array} { l l }
\frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \
0 & x

Question

Determine Whether the Function Is a Probability Density Function Over$$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \
0 & x < 0
\end{array} \right.$$

Examlex · Accepted Answer

The Answer of Determine Whether the Function Is a Probability Density Function Over$$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \
0 & x < 0
\end{array} \right.$$

Determine Whether the Function Is a Probability Density Function Over $f ( x ) = \left\{ \begin{array} { l l } \frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \\0 & x < 0\end{array} \right.$

Definitions:

Determine Whether the Function Is a Probability Density Function Over f(x)={4π(x2+1)x≥00x<0f ( x ) = \left\{ \begin{array} { l l } \frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \\0 & x < 0\end{array} \right.f(x)={π(x2+1)4​0​x≥0x<0​

Definitions:

Determine Whether the Function Is a Probability Density Function Over $f ( x ) = \left\{ \begin{array} { l l } \frac { 4 } { \pi \left( x ^ { 2 } + 1 \right) } & x \geq 0 \\0 & x < 0\end{array} \right.$