Let U be a continuous random variable with probability density function $$f ( x ) = \left\{ \begin{array} { l l }
\frac { 1 } { ( 1 - x ) ^ { 2 } } & 0 \leq x \leq .5 \
0 & \text { else. }
\end{array} \right.$$ Using the method of transformations, find the probability density function of $X = e ^ { U + 1 }$ .

Question

Let U be a continuous random variable with probability density function $$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 1 } { ( 1 - x ) ^ { 2 } } & 0 \leq x \leq .5 \
0 & \text { else. }
\end{array} \right.$$ Using the method of transformations, find the probability density function of $X = e ^ { U + 1 }$ .

Quizplus · Accepted Answer

The Answer of Let U be a continuous random variable with probability density function $$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 1 } { ( 1 - x ) ^ { 2 } } & 0 \leq x \leq .5 \
0 & \text { else. }
\end{array} \right.$$ Using the method of transformations, find the probability density function of $X = e ^ { U + 1 }$ .

Let U Be a Continuous Random Variable with Probability Density f(x)={1(1−x)20≤x≤.50 else. f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { ( 1 - x ) ^ { 2 } } & 0 \leq x \leq .5 \\0 & \text { else. }\end{array} \right.f(x)={(1−x)21​0​0≤x≤.5 else. ​

Let U Be a Continuous Random Variable with Probability Density $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { ( 1 - x ) ^ { 2 } } & 0 \leq x \leq .5 \\0 & \text { else. }\end{array} \right.$