FUNCTIONS OF A STANDARD NORMAL

The random variable X has a standard normal distribution. Find the PDF of the random variable Y, where:

1. Y=3X-1 , Y = 3X - 1

answer:

fY(y)=1/3*fX*(y+1/3)

f_ Y(y)=1/3*f_ X*(y+1/3)

2. Y=3X^2-1. For y>=-1,
Y = 3X^2 - 1. For y >= -1,

answer:

fY(y)=1/3*sqrt(3/y+1)*fX*sqrt(y+1/3)
f_Y(y)=1/3*sqrt(3/y+1)*f_X*sqrt(y+1/3)

(no need to answer. Public interest problem)

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To find the probability density function (PDF) of a random variable Y, where Y is a function of another random variable X with a standard normal distribution, we can use the method of transformation.

Let's go through each case:

1. Y = 3X - 1:

To find the PDF of Y, we need to find the cumulative distribution function (CDF) of Y and then differentiate it with respect to y to obtain the PDF.

First, let's find the CDF of Y:

F_Y(y) = P(Y ≤ y) = P(3X - 1 ≤ y) = P(X ≤ (y + 1)/3)

Since X has a standard normal distribution, we can look up the CDF of the standard normal distribution to find P(X ≤ (y + 1)/3). Let's denote the CDF of the standard normal distribution as Φ.

F_Y(y) = Φ((y + 1)/3)

To find the PDF f_Y(y), we differentiate the CDF with respect to y:

f_Y(y) = d/dy [Φ((y + 1)/3)]
= (1/3) * (1/√(2π)) * e^(-((y+1)/3)^2/2)

We can simplify this as:

f_Y(y) = (1/3) * (1/√(2π)) * e^(-((y+1)/3)^2/2)

f_Y(y) = (1/3) * f_X(y + 1/3)

where f_X(y) is the PDF of the standard normal distribution.

2. Y = 3X^2 - 1:

Similarly, we need to find the CDF of Y and differentiate it with respect to y to obtain the PDF.

F_Y(y) = P(Y ≤ y) = P(3X^2 - 1 ≤ y) = P(X ≤ √((y + 1)/3))

Since X has a standard normal distribution, we can look up the CDF of the standard normal distribution to find P(X ≤ √((y + 1)/3)). Again, let's denote the CDF of the standard normal distribution as Φ.

F_Y(y) = Φ(√((y + 1)/3))

To find the PDF f_Y(y), we differentiate the CDF with respect to y:

f_Y(y) = d/dy [Φ(√((y + 1)/3))]
= (1/2) * (1/√(6π)) * (√(1/3)) * (1/√((y + 1)/3)) * e^(-((y+1)/3)/2)

Simplifying this yields:

f_Y(y) = (1/6) * (1/√((y + 1)/3)) * (1/√(2π)) * e^(-((y+1)/3)/2)

f_Y(y) = (1/6) * √(3/(y + 1)) * f_X(√((y + 1)/3))

where f_X(y) is the PDF of the standard normal distribution.

Note: In both cases, f_X(y) refers to the probability density function (PDF) of the standard normal distribution, which is known and can be looked up in standard statistical tables or calculated using mathematical software.

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