The integral can be simplified by noting that the integrand is an even function of w, so we can integrate from 0 to ∞ and multiply by 2. We use a substitution u = w²(z²+1)/2, so du = w(z²+1)dw. Evaluating the definite integral yields the final PDF.

The integral can be simplified by noting that the integrand is an even function of w, so we can integrate from 0 to ∞ and multiply by 2. We use a substitution u = w²(z²+1)/2, so du = w(z²+1)dw. Evaluating the definite integral yields the final PDF.

If X and Y are independent standard normal random variables, find the probability density function of $Z = X/Y$ .

Answer: The probability density function of $Z = X/Y$ is $f_Z(z) = \frac{1}{\pi(z^2+1)}$ .

Step-by-step solution

Step 1: Define the joint PDF of X and Y

Since $X$ and $Y$ are independent standard normal random variables, their joint probability density function (PDF) is the product of their individual PDFs. The PDF of a standard normal variable is given by $f(t) = \frac{1}{\sqrt{2\pi}}e^{-t^2/2}$ .

Step 2: Use the transformation method

To find the PDF of $Z = X/Y$ , we use the transformation method. We introduce an auxiliary variable, say $W = Y$ , so that we have a one-to-one transformation from $(X, Y)$ to $(Z, W)$ . From $Z = X/Y$ , we can express $X$ in terms of $Z$ and $W$ as $X = ZW$ .

Step 3: Calculate the Jacobian of the transformation

The Jacobian of the transformation is needed to relate the joint PDF of $(X, Y)$ to the joint PDF of $(Z, W)$ . We calculate the partial derivatives of $x$ and $y$ with respect to $z$ and $w$ . Since $x = zw$ and $y = w$ , the Jacobian determinant is $w$ .

Step 4: Find the joint PDF of Z and W

The joint PDF of $Z$ and $W$ is obtained by substituting $x = zw$ and $y = w$ into $f_{X,Y}(x,y)$ and multiplying by the absolute value of the Jacobian. This gives us the joint density function in terms of $z$ and $w$ .

Step 5: Integrate out W to find the PDF of Z

To find the marginal PDF of $Z$ , we integrate the joint PDF $f_{Z,W}(z,w)$ with respect to $w$ over its entire range. Since $w = Y$ and $Y$ is a standard normal variable, $w$ can take any real value.

Step 6: Evaluate the integral

The integral can be simplified by noting that the integrand is an even function of $w$ , so we can integrate from $0$ to $\infty$ and multiply by $2$ . We use a substitution $u = w^2(z^2+1)/2$ , so $du = w(z^2+1)dw$ . Evaluating the definite integral yields the final PDF.

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If X and Y are independent standard normal random variables, find the probability density function of Z=X/YZ = X/YZ=X/Y.