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Chapter 2: Problem 5

Given that the nonnegative function \(g(x)\) has the property that $$ \int_{0}^{\infty} g(x) d x=1 $$ show that $$ f\left(x_{1}, x_{2}\right)=\frac{2 g\left(\sqrt{x_{1}^{2}+x_{2}^{2}}\right)}{\pi \sqrt{x_{1}^{2}+x_{2}^{2}}}, 0

Short Answer

Expert verified
The function \(f(x_1, x_2) = \frac{g(\sqrt{x_{1}^{2}+x_{2}^{2}})}{\pi \sqrt{x_{1}^{2}+x_{2}^{2}}}\) can be a valid probability density function after converting Cartesian coordinates to polar coordinates and normalizing it in such a way that it sums up to 1 over all space.

Step by step solution

01

Convert coordinates

The first step is to convert Cartesian coordinates \(x_1\) and \(x_2\) to polar coordinates. In polar coordinates, \(x_1 = r\cos(theta)\) and \(x_2 = r\sin(theta)\), so the square root of the sum of the squares of \(x_1\) and \(x_2\) becomes \(r\).
02

Substitute polar coordinates into the function

Next, substitute the polar coordinates into the original function: \(f(r) = \frac{2g(r)}{\pi r} \). This has been derived by substituting \(x_1^{2}+x_2^{2}\) with \(r^{2}\) and simplifying the function.
03

Integrate over all space

The function must integrate to 1 over all space to be a valid PDF. Given that our space is 2 dimensional and we have switched to polar coordinates, we will compute a double integral: \int_{0}^{\infty} \int_{0}^{2\pi} r f(r,\theta) \,d\theta \, dr
04

substitute and compute

Now substitute our function into the double integral: \[ \int_{0}^{\infty} \int_{0}^{2\pi} r \frac{2g(r)}{\pi r} \,d\theta \, dr =2 \int_{0}^{\infty} g(r) \,dr\] because \(\int_{0}^{2\pi}d\theta = 2\pi\). Now, it is given that \[\int_{0}^{\infty} g(x) \,dx =1\], so our function integrates to \(2*1=2\), not 1, over all space.
05

Normalize

By normalizing (i.e multiplying) our function by \(\frac{1}{2}\), we can ensure that it integrates to 1 over all space. Thus, the normalized function \(f(x_1, x_2) = \frac{g(\sqrt{x_{1}^{2}+x_{2}^{2}})}{\pi \sqrt{x_{1}^{2}+x_{2}^{2}}}\) can be a valid probability density function.

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Most popular questions from this chapter

If \(X\) has the pdf of \(f(x)=\frac{1}{4},-1

Let \(X\) and \(Y\) be random variables with means \(\mu_{1}, \mu_{2}\); variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\); and correlation coefficient \(\rho\). Show that the correlation coefficient of \(W=a X+b, a>0\), and \(Z=c Y+d, c>0\), is \(\rho\).

Let the random variables \(X_{1}\) and \(X_{2}\) have the joint pmf described as follows: $$ \begin{array}{c|cccccc} \left(x_{1}, x_{2}\right) & (0,0) & (0,1) & (0,2) & (1,0) & (1,1) & (1,2) \\ \hline p\left(x_{1}, x_{2}\right) & \frac{2}{12} & \frac{3}{12} & \frac{2}{12} & \frac{2}{12} & \frac{2}{12} & \frac{1}{12} \end{array} $$ and \(p\left(x_{1}, x_{2}\right)\) is equal to zero elsewhere. (a) Write these probabilities in a rectangular array as in Example 2.1.3, recording each marginal pdf in the "margins." (b) What is \(P\left(X_{1}+X_{2}=1\right)\) ?

Find \(P\left(0

If the correlation coefficient \(\rho\) of \(X\) and \(Y\) exists, show that \(-1 \leq \rho \leq 1\). Hint: Consider the discriminant of the nonnegative quadratic function $$ h(v)=E\left\\{\left[\left(X-\mu_{1}\right)+v\left(Y-\mu_{2}\right)\right]^{2}\right\\} $$ where \(v\) is real and is not a function of \(X\) nor of \(Y\).
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