Suppose that the random variables X and y are independent of each other and obey the uniform distribution on [0,1], the probability density of X + y is obtained

Suppose that the random variables X and y are independent of each other and obey the uniform distribution on [0,1], the probability density of X + y is obtained

10. Y is independent of each other and obeys the uniform distribution on [0,1] & nbsp; -- & gt; & nbsp; f (x, y) = 1
Z=X+Y
F (z) = P (x + Y & lt; Z) & nbsp; = & nbsp; ∫∫ f (x, y) DXDY & nbsp; = & nbsp; ∫∫ DXDY & nbsp; = area bounded by straight line x = 0, x = 1, y = 0, y = 1, y = - x + Z
When 0 & lt; Z & lt; 1, f (z) & nbsp; = & nbsp; (Z ^ 2) / 2
When 1 & lt; Z & lt; 2, f (z) & nbsp; = & nbsp; (Z ^ 2 / 2) - (Z-1) ^ 2
The probability density of Z = x + y
f(z) = dF(z)/dz=z      0<z<1;  f(z) = 2-z     1<z<2.
This is a step function. (U (x) (U (x) = 1, X & gt; 0, = 0, X & lt; 0, X & lt; 0, X & lt; 0 (U (x) (U (x) = 1, X (x) - U (Y-1) x, y is independent, z = x + y, so f (z) is the convolution of F (x) and f (x) - U (x) - U (x) - U (x) - U (x) - U (x) - U (x) - u (x) - U (U (x) (U (x) (U (x) (1, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, Z, Z, Z, Z, x, Z, x, Z, and Z, y, Z, and Z, and Z, and Z, f (f (Z, f (Z (Z (Z (Z (Z (Z f (z) = 0, the rest