Consider a random variable $U$ that has a uniform distribution on $(0,1)$ and a random variable $X$ that has a standard normal distribution. Assume that $U$ and $X$ are independent. Determine an expression for the probability density function of the random variable $Z = U + X $ in terms of the cumulative distribution function of $X$.

My attempt , $$f_Z(z)=\int_{u}f_U(u)f_{X}(z-u)du$$ $$=\int_{0}^{1}f_X(z-u)du$$



But the given answer is $$F_X(z)-F_X(z-1)$$


  • 3
    $\begingroup$ du=-dx. So you need a negative sign in the third line, or else reverse the limits. Note that your answer can’t be right, because it’s negative. $\endgroup$ Sep 6, 2018 at 3:53

1 Answer 1


You're making the substitution $x = z - u$ to transform the integral. The differential of this is:

$$ dx = 0 - du = - du $$

So the calculation finishes up like this:

$$=\int_{0}^{1}f_X(z-u)du = - \int_{z}^{z-1}f_X(x)dx = \int_{z-1}^{z}f_X(x)dx$$


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