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

Short Answer

Expert verified

Hence, the g(x) is proved in a decreasing function.

Step by step solution

01

Step:1 Given Information

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.

02

Step:2 Explanation of the solution

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

If , a<bthen

data-custom-editor="chemistry" g(b)<g(a)=f(a)<f(b)

If , a>bthen

data-custom-editor="chemistry" g(b)>g(a)=f(a)>f(b)

03

Step:3 Representation on graph

(b) Given

f(x)=x,g(x)=[x]+x-[x]

04

Step:4 Explanation of g(x) decreasing order

consider, f(x)=exand g(x)=-exTherefore it is evident that f(x) is an increasing function while g(x) is a decreasing function, as seen in the graph.


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