i would like to learn about this---Verjetnost, 2022/23, M. Perman, M. Raič 3. (20) Naj bodo X, Y in Z neodvisne standardne normalne slučajne spremenljivke. a. (10) Izračunajte gostoto slučajne spremenljivke W = $\sqrt{(X-Y)^2 + 2Z^2}$.

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In this problem, we need to find the density function of W, which is defined as the square root of the sum of (X minus Y) squared plus 2 times Z squared, where X, Y, and Z are independent standard normal random variables. Let's start by defining intermediate variables. We'll set U equal to X minus Y, and V equal to Z. Then W can be written as the square root of U squared plus 2 times V squared. Since X and Y are independent standard normal variables, U is a normal random variable with mean 0 and variance 2. The variable V is simply Z, which is a standard normal variable with mean 0 and variance 1. The blue curve shows the density of Z, while the red curve shows the density of U, which is more spread out due to its larger variance. Now, let's relate our variables to chi-squared distributions. Since U follows a normal distribution with mean 0 and variance 2, the random variable U-squared divided by 2 follows a chi-squared distribution with 1 degree of freedom. Similarly, since V is a standard normal variable, V-squared follows a chi-squared distribution with 1 degree of freedom. Let's define A as U-squared divided by 2, and B as V-squared. Both A and B are independent chi-squared variables with 1 degree of freedom. When we add two independent chi-squared variables, we get a new chi-squared variable whose degrees of freedom is the sum of the original degrees of freedom. So C equals A plus B follows a chi-squared distribution with 2 degrees of freedom. The density function of C is one-half times e to the negative c over 2 for c greater than zero. Finally, we can express W as the square root of 2 times C. Now we'll find the density function of W using the cumulative distribution function method. The CDF of W is the probability that W is less than or equal to some value w. Since W equals the square root of 2C, this is equivalent to the probability that the square root of 2C is less than or equal to w, which means 2C is less than or equal to w-squared, or C is less than or equal to w-squared over 2. We can express this as an integral of the density function of C from 0 to w-squared over 2. To find the PDF of W, we differentiate the CDF with respect to w. Using the fundamental theorem of calculus and the chain rule, we get one-half times e to the negative w-squared over 4, times the derivative of w-squared over 2, which is w. So the density function of W is w over 2 times e to the negative w-squared over 4 for w greater than 0, and 0 otherwise. The blue curve shows this density function, and the shaded area represents the cumulative distribution function up to a specific value of w. Let's examine some properties of the density function we derived. The density function of W is w over 2 times e to the negative w-squared over 4 for w greater than 0, and 0 otherwise. First, this function is always positive for positive values of w, which is a requirement for any probability density function. Second, as w approaches 0, the function also approaches 0. Third, as w approaches infinity, the exponential term dominates and the function approaches 0. Fourth, the function has a maximum at w equals 2, which we can find by taking the derivative and setting it equal to 0. We can verify that this is indeed a valid probability density function by checking that it integrates to 1 over its domain. Using the substitution u equals w-squared over 4, which means du equals w over 2 times dw, the integral becomes the integral of e to the negative u from 0 to infinity, which equals 1. This confirms that our density function is properly normalized. Let's summarize the key steps we took to find the density function of W. First, we defined intermediate variables U equals X minus Y and V equals Z. We identified that U follows a normal distribution with mean 0 and variance 2, while V follows a standard normal distribution with mean 0 and variance 1. We then transformed these variables to chi-squared distributions: U-squared divided by 2 follows a chi-squared distribution with 1 degree of freedom, and V-squared also follows a chi-squared distribution with 1 degree of freedom. Next, we summed these chi-squared variables to get C equals U-squared over 2 plus V-squared, which follows a chi-squared distribution with 2 degrees of freedom. We expressed W in terms of C as the square root of 2C. Finally, we applied the cumulative distribution function method and differentiated to find the probability density function. The resulting density function is w over 2 times e to the negative w-squared over 4 for w greater than 0, and 0 otherwise. This completes our solution to the problem.