4
$\begingroup$

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

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

$$=F_X(z-1)-F_X(z)$$

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

Why?

$\endgroup$
  • 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. $\endgroup$ – The Laconic Sep 6 '18 at 3:53
3
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.