1. fY(y)=1/5*fX(y+7 /5)
2. [[ fX(1+√(y+1)) + fX(1-√(y+1)) ]] / 2√(y+1)
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 ,
3 answers
Hey dz, would it be possible to get a more detailed explanation for you calculations?
Thanks!
Thanks!
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.
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.
3 answers