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if x, y, z are independent standard normal random variables, find the distribution of x²+y²+z². (hint: think spherical coordinates.)

User Vbroto
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The distribution of x²+y²+z², where x, y, and z are independent standard normal random variables, is a chi-squared distribution with 1 degree of freedom. The expression simplifies to r² in spherical coordinates.

To find the distribution of x²+y²+z², we can use spherical coordinates to simplify the expression. In spherical coordinates, x, y, and z can be expressed as:

x = r * sin(θ) * cos(φ)

y = r * sin(θ) * sin(φ)

z = r * cos(θ)

Here, r represents the radial distance, θ represents the polar angle, and φ represents the azimuthal angle.

Now, let's consider the expression x²+y²+z²:

(x²+y²+z²) = (r * sin(θ) * cos(φ))² + (r * sin(θ) * sin(φ))² + (r * cos(θ))²

Expanding the expression:

(x²+y²+z²) = r² * (sin²(θ) * cos²(φ) + sin²(θ) * sin²(φ) + cos²(θ))

Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can simplify the expression further:

(x²+y²+z²) = r² * (sin²(θ) * (cos²(φ) + sin²(φ)) + cos²(θ))

Since cos²(φ) + sin²(φ) = 1, the expression becomes:

(x²+y²+z²) = r² * (sin²(θ) + cos²(θ))

Using the trigonometric identity sin²(θ) + cos²(θ) = 1 again, we have:

(x²+y²+z²) = r²

This tells us that x²+y²+z² follows a chi-squared distribution with 1 degree of freedom.

User David Cummins
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