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

1. Y = 5X−7 .

2. Y = X2−2X . For y≥−1 ,

1. fY(y)=1/5*fX(y+7 /5)

2. [[ fX(1+√(y+1)) + fX(1-√(y+1)) ]] / 2√(y+1)

Sure, I'd be happy to explain in more detail!

1. To find the PDF of Y = 5X - 7, we can use the formula for transforming a continuous random variable.

fY(y) = fX(x) * |dx/dy|

Since Y = 5X - 7, we can solve for X in terms of Y to get:

X = (Y + 7)/5

Taking the derivative with respect to y:

dx/dy = 1/5

And substituting into the formula:

fY(y) = fX((y+7)/5) * 1/5

Since X has a standard normal distribution, we know that fX(x) is the standard normal probability density function:

fX(x) = (1/√(2π)) * e^(-x^2/2)

Substituting this into the formula above, we get:

fY(y) = (1/√(2π)) * e^(-((y+7)/5)^2/2) * 1/5

Simplifying:

fY(y) = (1/√(2π) * 5) * e^(-((y+7)/5)^2/2)

2. To find the PDF of Y = X^2 - 2X for y ≥ -1, we need to first find the range of X values that correspond to this range of Y values.

Y = X^2 - 2X can be rewritten as:

Y = (X-1)^2 - 1

Adding 1 to both sides:

Y + 1 = (X-1)^2

Taking the square root (note that since we want y ≥ -1, we take the absolute value of the square root):

|X-1| = √(Y+1)

Solving for X:

X = 1 + √(Y+1) or X = 1 - √(Y+1)

Now we can use the transformation formula again:

fY(y) = fX(x) * |dx/dy|

Since we have two solutions for X, we need to take the average of the two probabilities:

fY(y) = (fX(1+√(y+1)) + fX(1-√(y+1))) / 2 * |dx/dy|

Taking the derivative with respect to y:

dx/dy = 1/(2√(y+1))

Substituting everything we know into the formula:

fY(y) = (fX(1+√(y+1)) + fX(1-√(y+1))) / 2√(y+1)

Again, since X has a standard normal distribution, we know that fX(x) is the standard normal probability density function:

fX(x) = (1/√(2π)) * e^(-x^2/2)

Substituting this in, we get:

fY(y) = [(1/√(2π)) * e^(-(1+√(y+1))^2/2) + (1/√(2π)) * e^(-(1-√(y+1))^2/2)] / 2√(y+1)

Simplifying:

fY(y) = [[ fX(1+√(y+1)) + fX(1-√(y+1)) ]] / 2√(y+1)

I hope this helps clarify the calculations! Let me know if you have any further questions.

Hey dz, would it be possible to get a more detailed explanation for you calculations?

Thanks!

1. To find the PDF of Y = 5X - 7, we can use the transformation method. Let's start by finding the CDF of Y.

First, we express Y in terms of X:
Y = 5X - 7

Next, we find the inverse transformation to solve for X in terms of Y:
X = (Y + 7) / 5

Now, we differentiate this equation with respect to Y to find the PDF of Y:
dX/dY = 1/5

To get the PDF of Y, we need to multiply the derivative by the PDF of X. Since X has a standard normal distribution, its PDF is given by:
f_X(x) = (1/√(2π)) * e^(-x^2/2)

Therefore, the PDF of Y can be found using the equation for the PDF of a transformed random variable:
f_Y(y) = f_X((y + 7) / 5) * (dX/dY)
= (1/√(2π)) * e^(-((y + 7) / 5)^2/2) * (1/5)
= (1/(5√(2π))) * e^(-(y + 7)^2/50)

So, the PDF of Y is given by:
f_Y(y) = (1/(5√(2π))) * e^(-(y + 7)^2/50)

2. To find the PDF of Y = X^2 - 2X, where y ≥ -1, we can again use the transformation method. However, in this case, we need to consider the given restriction, y ≥ -1.

First, we express Y in terms of X:
Y = X^2 - 2X

Next, we complete the square to rewrite Y in terms of a quadratic form:
Y = (X^2 - 2X + 1) - 1
= (X - 1)^2 - 1

Since Y ≥ -1, this implies that (X - 1)^2 - 1 ≥ -1. Simplifying this inequality, we get (X - 1)^2 ≥ 0, which is always true. Therefore, we don't need to consider any restrictions when finding the PDF of Y.

To find the PDF of Y, we differentiate the inverse transformation equation, X = 1 ± √(Y + 1), with respect to Y:
dX/dY = ± (1/2) * (1/√(Y + 1))

Since X has a standard normal distribution, its PDF is given by:
f_X(x) = (1/√(2π)) * e^(-x^2/2)

Therefore, the PDF of Y can be found using the equation for the PDF of a transformed random variable:
f_Y(y) = f_X(1 ± √(y + 1)) * |(dX/dY)|
= (1/√(2π)) * e^(-(1 ± √(y + 1))^2/2) * (1/2) * (1/√(y + 1))

Simplifying this expression gives:
f_Y(y) = (1/(2√(2π))) * e^(-(1 ± √(y + 1))^2/2) * (1/√(y + 1))

Therefore, the PDF of Y is given by:
f_Y(y) = (1/(2√(2π))) * e^(-(1 ± √(y + 1))^2/2) * (1/√(y + 1))

Remember to use the appropriate sign in the ± based on if the quadratic form in the transformation equation, (X - 1)^2 - 1, is positive or negative.

To find the probability density function (PDF) of a random variable Y, given a transformation of another random variable X, we can use the method of transformations.

1. Y = 5X - 7:

The first step is to find the cumulative distribution function (CDF) of Y. We'll then differentiate the CDF to get the PDF.

CDF of Y:
F_Y(y) = P(Y ≤ y) = P(5X - 7 ≤ y)
= P(X ≤ (y + 7)/5)
= F_X((y + 7)/5) (by substituting the CDF of X)

Next, we differentiate the CDF of Y to obtain the PDF of Y:

PDF of Y:
f_Y(y) = d/dy [F_Y(y)]
= d/dy [F_X((y + 7)/5)]
= (1/5) * f_X((y + 7)/5) (by using the chain rule)

Since X has a standard normal distribution, its PDF is given by:

f_X(x) = (1/√(2π)) * e^(-x^2/2)

So, substituting this into the expression for f_Y(y), we have:

f_Y(y) = (1/5) * f_X((y + 7)/5)
= (1/5) * (1/√(2π)) * e^(-((y + 7)/5)^2/2)

Thus, the PDF of Y = 5X - 7 is given by:

f_Y(y) = (1/5√(2π)) * e^(-((y + 7)/5)^2/2)

2. Y = X^2 - 2X for y ≥ -1:

Following the same steps, we'll find the CDF of Y and then differentiate it to obtain the PDF.

CDF of Y:
F_Y(y) = P(Y ≤ y) = P(X^2 - 2X ≤ y)
= P(X^2 - 2X - y ≤ 0)
= P((X - 1)^2 - 1 - y ≤ 0)
= P((X - 1)^2 ≤ 1 + y)
= P(-√(1 + y) ≤ X - 1 ≤ √(1 + y))
= P(1 - √(1 + y) ≤ X ≤ 1 + √(1 + y))
= F_X(1 + √(1 + y)) - F_X(1 - √(1 + y))

Again, we differentiate the CDF of Y to get the PDF:

PDF of Y:
f_Y(y) = d/dy [F_Y(y)]
= d/dy [F_X(1 + √(1 + y)) - F_X(1 - √(1 + y))]
= d/dy [F_X(1 + √(1 + y))] - d/dy [F_X(1 - √(1 + y))]
= f_X(1 + √(1 + y)) * (1/2√(1 + y)) - f_X(1 - √(1 + y)) * (-1/2√(1 + y)) (using the chain rule)

Substituting the PDF of X into the expression for f_Y(y), we have:

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

Thus, the PDF of Y = X^2 - 2X for y ≥ -1 is given by:

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