let x be a continuous random variable variable with probability distribution f(y)=3y, 0 1<=y<=2. find the probability distribution of the random variables Y=4y-1
To find the probability distribution of the random variable Y=4y-1, we need to calculate the cumulative distribution function (CDF) of Y and then differentiate it to find the probability density function (PDF).
Let's start by finding the CDF of Y.
CDF of Y = P(Y ≤ y)
= P(4y-1 ≤ y)
= P(4y ≤ y+1)
= P(y ≤ (y+1)/4)
= F((y+1)/4) (F is the CDF of x)
Since the probability distribution of x is given by f(y) = 3y, where 1 ≤ y ≤ 2, we can find the CDF of x by integrating the density function.
CDF of x = ∫(3y) dy from 1 to y (limits of integration from 1 to y)
= (3/2)y^2 - (3/2)
= 3/2(y^2 - 1)
Now substituting y with (y+1)/4 in the CDF of x to get the CDF of Y:
CDF of Y = F((y+1)/4) = 3/2[((y+1)/4)^2 - 1]
Next, we differentiate the CDF of Y with respect to y to find the PDF of Y.
PDF of Y = d/dy (CDF of Y)
= d/dy [3/2[((y+1)/4)^2 - 1]]
= 3/2 * d/dy[((y+1)/4)^2] - 3/2 * d/dy [1]
= 3/2 * 2 * ((y+1)/4) * (1/4) - 0
= 3/16 * (y+1)
Therefore, the probability distribution of the random variable Y=4y-1 is f(y) = 3/16 * (y+1) for 3 ≤ y ≤ 7. (Obtained by substituting the range of y values into the derived PDF of Y.)