# Probability Density Function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1)

• forumfann
In summary, the probability density function (pdf) of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) can be found by first determining the distribution of |X|^(1/2), using a standard transformation, and then taking into account the independent and identical distribution of X, Y, and Z.
forumfann
Could anyone help me figure out the the probability density function (pdf) of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) if X, Y and Z are distributed normally with mean 0 and variance 1, N(0,1) ?

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Are your variables independent? If so, first work out the distribution of $$|X|^{1/2}$$. The other distributions will be identical and you can use standard procedures to find the distribution of their sum.

Yes, the variables are independent. But what are the standard procedures? It there a easier way to get the pdf if one has more random variables than three?

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If X, Y, and Z, are distributed normally, just make them a normal (gausian) PDF of a random variable.

X=Ae^(ax). You're lucky that you can use little x,y, and z for your random variables to match up with their distributions.

"X=Ae^(ax). " That isn't the form of a normal distribution.

Okay, suppose $$X \sim n(0,1)$$. Think this way.

1) You should be able to write down the distribution of $$|X|$$ - it's a pretty
standard result, and if you're working on this problem I'm guessing you know this.
2) Use a standard transformation (if $$W = |X|, find the distribution of square root of W). This gives the distribution of [tex] |X|^{1/2}$$.
3) Since $$X, Y$$ and $$Z$$ are i.i.d, the same is true for

$$|X|^{1/2} + |Y|^{1/2} + |Z|^{1/2}$$

so the distribution of their sum should be relatively easy to obtain.

## 1. What is a probability density function?

A probability density function is a mathematical function that describes the probability distribution of a continuous random variable. It gives the likelihood of a random variable taking on a specific value or falling within a certain range of values.

## 2. What does N(0,1) refer to in the probability density function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1)?

N(0,1) refers to a normal distribution with a mean of 0 and a standard deviation of 1. This is a common distribution used in statistics to model random variables.

## 3. What is the significance of the absolute value in the probability density function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1)?

The absolute value ensures that the values of the random variables X, Y, and Z are always positive. This is important because the square root of a negative number is not a real number, so the absolute value ensures that the function is well-defined and can be integrated.

## 4. How can the probability density function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1) be used in practical applications?

The probability density function can be used to calculate the probability of a certain event occurring based on the values of X, Y, and Z. This can be useful in various fields such as finance, engineering, and physics to model and analyze data.

## 5. Is the probability density function of |X|^(1/2)+|Y|^(1/2)+|Z|^(1/2) with N(0,1) symmetric?

Yes, the probability density function is symmetric because the absolute value function ensures that the values on either side of the mean are reflected and have equal probabilities. This is a characteristic of the normal distribution.

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