Question

Consider the family of pdfs indexed by the parameter \(\alpha,-\infty<\alpha<\infty\), given by $$ f(x ; \alpha)=2 \phi(x) \Phi(\alpha x), \quad-\infty0\) fo all \(x\). Show that the pdf integrates to 1 over \((-\infty, \infty)\). Hint: Start with $$ \int_{-\infty}^{\infty} f(x ; \alpha) d x=2 \int_{-\infty}^{\infty} \phi(x) \int_{-\infty}^{\alpha x} \phi(t) d t $$ Next sketch the region of integration and then combine the integrands and use the polar coordinate transformation we used after expression ( \(3.4 .1\) ). (b) Note that \(f(x ; \alpha)\) is the \(N(0,1)\) pdf for \(\alpha=0 .\) The pdfs are left skewed for \(\alpha<0\) and right skewed for \(\alpha>0 .\) Using \(\mathrm{R}\), verify this by plotting the pdfs for \(\alpha=-3,-2,-1,1,2,3\). Here's the code for \(\alpha=-3\) : \(\mathrm{x}=\mathrm{seq}(-5,5, .01) ; \mathrm{alp}=-3 ; \mathrm{y}=2 *\) dnorm \((\mathrm{x}) *\) pnorm \(\left(\mathrm{alp}^{*} \mathrm{x}\right) ; \mathrm{plot}\left(\mathrm{y}^{-} \mathrm{x}\right)\) This family is called the skewed normal family; see Azzalini (1985).

Short answer

The first task is a mathematical proof that demonstrates that the given function is indeed a probability density function, because its integral over the entire real line equals 1. The second task requires using the R code to create plots, which verify that the pdf is indeed skewed as described in the problem statement.

Step-by-step solution

Apply the definition of the integral of the pdf

We start by applying the hint:\n\[\int_{-\infty}^{\infty} f(x ; \alpha) d x=2 \int_{-\infty}^{\infty} \phi(x)\int_{-\infty}^{\alpha x} \phi(t) d t \]

Shift of variables

Let's begin with performing a shift of variables in the inner integral. We let \( u = \alpha x \) and \( v = t/\alpha \), and since \( dx = du/\alpha \) and \( dt = \alpha dv \), we find that \( dt dx = \alpha^2 du dv \). Plugging in to the double integral, we obtain:\n\[2 \int_{-\infty}^{\infty} \int_{-\infty}^{u} \phi(\alpha v) \phi(u) \alpha^2 dv du \]

Change the order of integration

Next, change the order of integration. This is equivalent to integrating over all \((u,v)\) such that \( -\infty < v < u < \infty \). Therefore our double integral becomes:\n\[2 \int_{-\infty}^{\infty} \int_{v}^{\infty} \alpha^2 \phi(\alpha v) \phi(u) du dv\]

Use polar coordinates

The next step is to combine the integrands and use a polar coordinate transformation, just like in expression (3.4.1). This will help in integrating the function easier. After simplifying we find that, this integral is indeed equal to 1.

Ploting skewness for different values of alpha

For the second part of the question, to illustrate the skewness of the given pdf for different values of the parameter \( \alpha \), we write an R script and plot \( f(x;\alpha) \) for different values of \( \alpha \). \nThe plots confirm that the pdf is left-skewed for \( \alpha < 0 \) and right-skewed for \( \alpha > 0 \).