Question

Let X and Y be iid uniform random variables on [0,1]. Find the pdf of Z=X+Y

Answer 1

Answer 2

To find the probability density function (PDF) of the random variable Z = X + Y, where X and Y are independent and identically distributed (iid) uniform random variables on the interval [0,1], we can use the concept of convolution.

The convolution of two continuous random variables can be found by integrating their joint probability density function. Since X and Y are independent, their joint probability density function is the product of their individual probability density functions.

The probability density function of a uniform random variable on the interval [a, b] is given by:

f_X(x) = 1 / (b - a), for a ≤ x ≤ b f_Y(y) = 1 / (b - a), for a ≤ y ≤ b

Since X and Y are both uniform on [0,1], their individual PDFs are:

f_X(x) = 1, for 0 ≤ x ≤ 1 f_Y(y) = 1, for 0 ≤ y ≤ 1

Now, to find the PDF of Z = X + Y, we need to determine the range of Z. Since both X and Y are between 0 and 1, the minimum value of Z is 0 (when both X and Y are 0), and the maximum value of Z is 2 (when both X and Y are 1).

For 0 ≤ z ≤ 2, the PDF of Z is given by the convolution:

f_Z(z) = ∫[0,1] f_X(x) * f_Y(z - x) dx

Substituting the PDFs of X and Y, we have:

f_Z(z) = ∫[0,1] 1 * 1 dz f_Z(z) = ∫[0,1] dz f_Z(z) = z, for 0 ≤ z ≤ 1

Since the PDF must integrate to 1 over its entire range, we also need to consider the PDF for 1 < z ≤ 2. In this range, Z can only be obtained when X + Y > 1. To find the PDF in this range, we integrate over the region where X + Y > 1:

f_Z(z) = ∫[z-1,1] f_X(x) * f_Y(z - x) dx

Substituting the PDFs of X and Y, we have:

f_Z(z) = ∫[z-1,1] 1 * 1 dx f_Z(z) = ∫[z-1,1] dx f_Z(z) = 2 - z, for 1 < z ≤ 2

Therefore, the PDF of Z is given by:

f_Z(z) = z, for 0 ≤ z ≤ 1 f_Z(z) = 2 - z, for 1 < z ≤ 2 0, otherwise

This completes the PDF of Z = X + Y for 0 ≤ z ≤ 2. Outside this range, the PDF is zero since the sum of two random variables cannot exceed 2 in this case.