Question

In Example 3.8.4, the p.d.f. of \({\bf{Y = }}{{\bf{X}}^{\bf{2}}}\) is much larger for values of y near 0 than for values of y near 1 despite the fact that the p.d.f. of X is flat. Give an intuitive reason why this occurs in this example.

Short answer

There is an inverse relationship between the rise in y and the function value.

Step-by-step solution

Given information

The pdf mentioned in example 3.8.4 is a uniform distribution\(X \sim U\left[ { - 1,1} \right]\). Therefore the p.d.f. is:

\(\begin{aligned}f\left( x \right) &= \frac{1}{2}, - 1 \le x \le 1,\\ &= 0,otherwise\end{aligned}\)

Given figure is:

CDF and PDF of Y

Let us define \(Y = {X^2}\)

What is the CDF approach?

A CDF approach is a method of random variable transformation wherein the pdf of the new variable is fetched from the CDF of the new variable, which is in terms of the CDF of the old variable.

The CDF approach steps,

• We substitute the Y variable in the CDF formula in a CDF approach.
• We then substitute Y in terms of X.
• We reduce this form until we bring the CDF in terms of X.
• Since we have already calculated the CDF of X, we replace the form of variable y in the formula for X.
• In the final step, we get the CDF of Y variable as an expression of the CDF of X with y variables.

By using the CDF approach.

\(\begin{aligned}G\left( y \right) &= P\left( {Y \le y} \right)\\ &= P\left( {{X^2} \le y} \right)\\ &= P\left( { - {y^{\frac{1}{2}}} \le X \le {y^{\frac{1}{2}}}} \right)\\ &= \int_{ - {y^{\frac{1}{2}}}}^{{y^{\frac{1}{2}}}} {f\left( x \right)} dx\\ &= {y^{^{\frac{1}{2}}}}\end{aligned}\)

Therefore, the CDF of Y is \({y^{^{\frac{1}{2}}}}\)

Convert the CDF into PDF

For 0<y<1, it follows that p.d.f g(y) of Y is

\(\begin{aligned}g\left( y \right) &= \frac{{dG\left( y \right)}}{{dy}}\\ &= \frac{1}{{2{y^{\frac{1}{2}}}}}\end{aligned}\)

Reasoning for the diagram

The range \(Y = {X^2}\)has r range [0,1].All the values are therefore positive.

Y is simply the square of a random variable with a uniform distribution, the p.d.f. of Y is unbounded in the neighborhood of y = 0.

The pdf of Y has the variable as the denominator. Therefore, there is an inverse relation. With the rise in the value of y, the value of function falls