# Finding PDF of Y if X is a Gaussian PDF

• Satwant
In summary: Y, you only need to find the density for y > 0, perhaps using the Fundamental Theorem of Calculus. In summary, To find the PDF of Y, a random variable where Y = |X| and X has a Gaussian PDF, we can use the CDF of X, Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2), and replace X^2 with Y. This gives us the CDF of Y, G(y) = 2*Fi(\sqrt{y}) - 1. Then, we can take the derivative of G(y) to find the density of Y, which is
Satwant
Find the PDF of Y if X is a Gaussian PDF:
fx(x) = (1e-x^2/2)/(2pi)^1/2 ; -infnity<x<+infinity

Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2)

b) Sketch both the PDF, fY(y) and CDF, FY(y) for randon variable Y

Satwant said:
Find the PDF of Y if X is a Gaussian PDF:
fx(x) = (1e-x^2/2)/(2pi)^1/2 ; -infnity<x<+infinity

Fi(x) = Integral -infnty to + infnity((1e-x^2/2)/(2pi)^1/2)

b) Sketch both the PDF, fY(y) and CDF, FY(y) for randon variable Y

You haven't given a relationship between X and Y have you?

Sorry, it is Y = X^2

Then just replace $X^2$ in the formulas by Y!
$$fx(Y) = e^{-Y/2}/(2\pi)^{1/2}$$
$$0\le Y< \infty$$

$$Fi(x) = \int_{-\infty}^\infty (e^{-Y/2}/(2\pi)^{1/2})dy$$

but if Y = modX

Also, to find PDF of Y of a Gaussian PDF, do I need to find mean and variance?

xx yyy

I don't know what you mean by "modX" or by "find PDF of Y of a Gaussian PDF".

I mean
Y = |X|

you told me how to do it for Y = X^2 but in another part, how to do it for
Y = |X|

Try this. For the distribution function of $$X$$ write

$$F(x) = \Pr(X \le x) = \Phi(x)$$

Here $$\Phi(x)$$ is the usual notation for the CDF of the standard Gaussian.
I'll use $$G(y)$$ as the CDF for your new random variable.

\begin{align*} G(y) & = \Pr(Y \le y) = P(X^2 \le y) \\ & =\Pr(-\sqrt{y} \le X \le \sqrt{y}) \\ & = \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) \end{align*}

Since the standard Gaussian is symmetric around 0,

$$\Phi(-a) = 1 - \Phi(a)$$

for any number $$a$$. From the place where I left off:

\begin{align*} G(y) &= \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = \Phi(\sqrt{y}) - (1- \Phi(\sqrt{y}))\\ & = 2\Phi(\sqrt{y}) - 1 \end{align*}

Now use these facts:

* The density of $$Y$$ is the derivative of $$G(y)$$
* You need to use the chain rule when you take the derivative of $$\Phi(\sqrt{y})$$
* The derivative of $$\Phi(x)$$ is the density of the standard Gaussian
* The random variable $$Y$$ is defined on $$(0, \infty)$$
* The distribution of $$Y$$ is one you should be able to recognize

the method for your second question is similar:
$$\Pr(|X| \le y) = \Pr(-y \le X \le y)$$

go from here. (I have a second $$\le$$ between X and y above, but it isn't showing.)

thanks!

Any idea how to sketch the PDF, fY(y) and CDF, FY(y) for randon variable Y for this problem?

Thank you.

once you have the expression for the density - graph it as any other function.
You'll need a technology aid to graph the CDF.

When finding the density of Y by taking the derivative of G(y) as described earlier, we want to take the derivative of G(y) in respect to y correct? Or is it in respect to phi?

Thank you for all the help

$$g(y) = \frac{dG}{dy}$$

Thanks

## 1. What is a Gaussian PDF?

A Gaussian PDF, also known as a normal distribution, is a bell-shaped probability distribution that is commonly used to model continuous random variables in statistics. It is characterized by its mean and standard deviation, and is widely used in many applications due to its simplicity and versatility.

## 2. What is the relationship between X and Y in a Gaussian PDF?

In a Gaussian PDF, Y is a random variable that is dependent on the random variable X. This means that the value of Y is determined by the value of X, and the relationship between the two can be described by the parameters of the normal distribution.

## 3. How do you find the PDF of Y if X is a Gaussian PDF?

To find the PDF of Y if X is a Gaussian PDF, you can use a mathematical formula known as the transformation rule. This rule states that if Y = g(X), where g is a function, then the PDF of Y can be calculated by taking the derivative of the PDF of X with respect to Y and then multiplying it by the absolute value of the derivative of g with respect to X.

## 4. Can you provide an example of finding the PDF of Y if X is a Gaussian PDF?

Sure! Let's say X is a Gaussian PDF with a mean of 0 and a standard deviation of 1. If we define Y = 2X + 1, then the PDF of Y can be calculated as follows:

PDF(Y) = |2| * PDF(X) * (1/2) = 2 * (1/sqrt(2*pi)) * e^(-1/2 * ((y-1)/2)^2)

This formula can be used to find the PDF of Y for any given value of X and any function g that relates X and Y.

## 5. What are some important properties of a Gaussian PDF?

Some important properties of a Gaussian PDF include:

• The mean, median, and mode are all equal and located at the center of the distribution.
• The distribution is symmetric, with 50% of the data falling on either side of the mean.
• About 68% of the data falls within one standard deviation of the mean, and about 95% falls within two standard deviations.
• The tails of the distribution approach but never touch the x-axis, meaning that there is a small but non-zero probability of extreme values occurring.
• Many natural phenomena tend to follow a normal distribution, making it a useful tool in data analysis and modeling.

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