1
$\begingroup$

Suppose we have two random variables $A$ and $B$ both uniformly distributed on $[0, 1]$. Define $Z=A+B$.

The moment generating function of $A$ and $B$ is $M_A(\theta)=M_B(\theta)=\frac{e^{\theta}-1}{\theta}$, so $M_Z(\theta)=M_A(\theta) M_B(\theta)=(\frac{e^{\theta}-1}{\theta})^2 = \frac{e^{2\theta}-2e^{\theta}+1}{\theta^2} $.

On the other hand the probability density function of $Z$ is $f_Z(z)=z$ for $z<1$ so we should have $M_Z(\theta)=E[e^{Z\theta}]=\int_0^1e^{z\theta}f_Z(z) \space dz= \int_0^1e^{z\theta}z \space dz = \frac{e^{\theta} \theta - e^{\theta} + 1}{\theta^2}$.

Can someone please explain me where does the difference of answers come from?

$\endgroup$
3
$\begingroup$

"On the other hand the probability density function of $Z$ is $f_Z(z) = z$ for $z<1$, so [...]

Why would that be the case? $Z$ is not uniform, it has a triangular distribution with parameters $a=0$, $b=2$ and $c=1$. You can check the MGF matches.

To get back the result with your computations: $f_Z$ is actually $$ f_Z(z) = z\mathbb{1}_{[0,1]}(z) - (2-z)\mathbb{1}_{[1,2]}(z) $$ (note that your proposed expression would integrate to $1/2$, not to $1$!). You can check that $$ \int_{0}^1 e^{z\theta}z \, dz + \int_{1}^2 e^{z\theta}(2-z) \, dz = \frac{(e^\theta-1)^2}{\theta^2} $$ as it should.

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/357672/… what about this, then? $\endgroup$ – harlem May 4 '18 at 20:54
  • 2
    $\begingroup$ @harlem The link is correct (see also my edited answer), your expression is not. The support of the PDF is $[0,2]$, with one expression for $[0,1]$ and another for $[1,2]$. You only kept the first part... $\endgroup$ – Clement C. May 4 '18 at 20:56

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.