Probability density: change of variable

[longrob]

If x is a random variable uniformly continuously distributed on [0.1], and y=x^3, then y has the density:

$$\frac{1}{3}y^{-2/3}$$

on [0,1]

But, if x has the same distribution, but on [-0.5, 0.5], there seems to be a problem because we have $$y^{-2/3}$$ for negative values of y. This is overcome if we use the absolute value of y, so in this case we get:

$$\frac{1}{3}\mid y\mid{}^{-2/3}$$

on [-1/8, 1/8]

Is this correct ? It seems to be, since integrating it yields 1, but how can I justify just replacing y with |y| ?

 

[mmwave]

It is important to note that the density function \frac{1}{3}y^{-2/3} on [0,1] is only valid for positive values of y. So when we are dealing with a random variable that can take on negative values, such as in the case of x on [-0.5, 0.5], we need to adjust the density function accordingly.

Using the absolute value of y, as in \frac{1}{3}\mid y\mid{}^{-2/3} on [-1/8, 1/8], is a valid way to adjust the density function. This is because the absolute value function essentially "flips" negative values to positive values, so we are still using the same density function for positive values of y.

To justify this replacement, we can look at the properties of the absolute value function. The absolute value function is continuous and differentiable everywhere except at 0. Since our density function is also continuous and differentiable everywhere except at 0, we can safely use the absolute value function without affecting the overall properties of the density function.

Additionally, integrating the adjusted density function over the appropriate range, in this case [-1/8, 1/8], still yields 1, which is the desired result for a valid density function.

In conclusion, using the absolute value of y in the density function \frac{1}{3}\mid y\mid{}^{-2/3} on [-1/8, 1/8] is a valid way to handle the problem of negative values of y, and can be justified through the properties of the absolute value function.