Question

Let X have the uniform distribution on the interval, and let prove that \({\bf{cX + d}}\) it has a uniform distribution on the interval \(\left[ {{\bf{ca + d,cb + d}}} \right]\)

Short answer

Therefore, the following theorem is established: If X is a random variable following uniform distribution on a given interval (a,b) that is, then \(cX + d \sim U\left( {ca + d,cb + d} \right)\)

Step-by-step solution

Given information

X is a random variable following uniform distribution on a given interval (a,b) that is, \(X \sim U\left[ {a,b} \right]\)

Define pdf of X

\(\begin{aligned}{f_x} &= \frac{1}{{\left( {b - a} \right)}},a < x < b\\ &= 0,otherwise\end{aligned}\)

Using Linear Transformation Proof

According to theorem 3.8.2, Suppose that X is a random variable for which the p.d.f. is\(f\)and that\(Y = aX + b(a \ne 0).\) Then the p.d.f of Y is

\(\begin{aligned}g\left( y \right) &= \frac{1}{{\left| a \right|}}f\left( {\frac{{y - b}}{a}} \right), - \infty < y < \infty \\ &= 0,otherwise\end{aligned}\)

Step 4:Substituting the values in Linear Transformation

Let us define a new variable \({\bf{Y = cX + d}}\).

The p.d.f of Y is

\(\)\(\begin{aligned}g\left( y \right) &= \frac{1}{{\left| c \right|}}f\left( {\frac{{y - d}}{c}} \right), - \infty < y < \infty \\ &= 0,otherwise\\ &= \frac{1}{{\left| c \right|}} \times \frac{1}{{\left( {b - a} \right)}}\end{aligned}\)

Therefore, the pdf of \(Y \sim U\left[ {ca + d,cb + d} \right]\)

The above result directly applies the theorem, and hence, this linear transformation stored the uniform distribution.

Therefore, the pdf follows a uniform distribution.\( \sim U\left( {ca + d,cb + d} \right)\)