Prove Theorem 7.1 when g(x) is a decreasing function.

Let X be a continuous random variable with the probability density function f(X) as its probability density function. Assume that g(x) is a differentiable (and so continuous) function of x that is strictly monotonic (growing or decreasing). Let X be a continuous random variable with probability density function f. Then the random variable Y defined by Y = g(X) has a probability density function provided by Let X be a continuous random variable with probability density function f. (X). Assume that g(x) is a strictly monotonic (growing or decreasing), differentiable (and) differentiable (and) differentiable (and) differentiable (and) differentiable (and) differentiable (and) differentiable

As a result, the function of x is continuous. Then there's the random variable Y, which is defined as Y = g. (X)

A probability density function is provided.

(a) Assume that f is increasing while g is decreasing, and that f(a)=g (a)

If , $a<b$then

data-custom-editor="chemistry" $\mathrm{g}\left(\mathrm{b}\right)<\mathrm{g}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{a}\right)<\mathrm{f}\left(\mathrm{b}\right)$

If , $a>b$then

data-custom-editor="chemistry" $\mathrm{g}\left(\mathrm{b}\right)>\mathrm{g}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{a}\right)>\mathrm{f}\left(\mathrm{b}\right)$

(b) Given

$f\left(x\right)=x,g\left(x\right)=\left[x\right]+\sqrt{x-\left[x\right]}$

consider, $f\left(x\right)=e^x$and $g\left(x\right)=-e^x$Therefore it is evident that f(x) is an increasing function while g(x) is a decreasing function, as seen in the graph.