Confusion about determining distribution of sum of two random variables

[psie]

Homework Statement

Let $X$ and $Y$ be independent r.v. such that $X\in U(0,1)$ and $Y\in U(0,\alpha)$. Find the density function of $Z=X+Y$. Remark: Note that there are two cases: $\alpha\geq 1$ and $\alpha <1$.

Relevant Equations

The relevant equation is that the pdf of the sum of two continuous random variables is a convolution.

Let's recall the densities of $X$ and $Y$:
\begin{align}
f_X(x)=\mathbf{1}_{(0,1)}(x), \quad f_Y(y)=\frac{1}{\alpha}\mathbf{1}_{(0,\alpha)}(y)
\end{align}
Let $z\in (0,1+\alpha)$. So we know that $f_Z(z)$ is given by:
\begin{align}
f_Z(z)=\int_\mathbb{R} f_X(t)f_Y(z-t)\,dt
\end{align}
Both $f_X$ and $f_Y$ are zero most of the time. We check $f_X$ and $f_Y$ one by one. We start with $f_X(t)$; it is nonzero when $0<t<1$. We have that $f_Y(z-t)$ is nonzero when $0<z-t<\alpha$. That means $t<z$ and $z-\alpha<t$. We want to satisfy all these inequality at once. So that means $\max\{z-\alpha,0\}<t<\min\{1,z\}$. Hence:
\begin{align*}
f_Z(z)=\int_{\mathbb{R}}f_X(t)f_Y(z-t)\,dt=\int^{\min\{1,z\}}_{\max\{z-\alpha,0\}}\frac{1}{\alpha}\,dt=\frac{\min\{1,z\}- \max\{z-\alpha,0\}}{\alpha}
\end{align*}

Now, what troubles me is the remark in the problem statement. I don't see that there are two cases to consider. For me, the density is simply the one given in the last equation, or?

 

[Orodruin]

Consider the case $\alpha = 1$ and $z = -1$. Your function would be
$$
f_Z(-1) = \min(1,-1) - \max(-1-1,0) = -1 - 0 = -1.
$$
Is this reasonable?

 

[WWGD]

Homework Statement: Let $X$ and $Y$ be independent r.v. such that $X\in U(0,1)$ and $Y\in U(0,\alpha)$. Find the density function of $Z=X+Y$. Remark: Note that there are two cases: $\alpha\geq 1$ and $\alpha <1$.
Relevant Equations: The relevant equation is that the pdf of the sum of two continuous random variables is a convolution.

Let's recall the densities of $X$ and $Y$:
\begin{align}
f_X(x)=\mathbf{1}_{(0,1)}(x), \quad f_Y(y)=\frac{1}{\alpha}\mathbf{1}_{(0,\alpha)}(y)
\end{align}
Let $z\in (0,1+\alpha)$. So we know that $f_Z(z)$ is given by:
\begin{align}
f_Z(z)=\int_\mathbb{R} f_X(t)f_Y(z-t)\,dt
\end{align}
Both $f_X$ and $f_Y$ are zero most of the time. We check $f_X$ and $f_Y$ one by one. We start with $f_X(t)$; it is nonzero when $0<t<1$. We have that $f_Y(z-t)$ is nonzero when $0<z-t<\alpha$. That means $t<z$ and $z-\alpha<t$. We want to satisfy all these inequality at once. So that means $\max\{z-\alpha,0\}<t<\min\{1,z\}$. Hence:
\begin{align*}
f_Z(z)=\int_{\mathbb{R}}f_X(t)f_Y(z-t)\,dt=\int^{\min\{1,z\}}_{\max\{z-\alpha,0\}}\frac{1}{\alpha}\,dt=\frac{\min\{1,z\}- \max\{z-\alpha,0\}}{\alpha}
\end{align*}

Now, what troubles me is the remark in the problem statement. I don't see that there are two cases to consider. For me, the density is simply the one given in the last equation, or?

Nitpick: Convolutionof _Independent_ Random Variables.

 

[psie]

Consider the case $\alpha = 1$ and $z = -1$. Your function would be
$$
f_Z(-1) = \min(1,-1) - \max(-1-1,0) = -1 - 0 = -1.
$$
Is this reasonable?

How can $z=-1$? If $\alpha=1$, then $z\in (0,2)$, no?

 

[Orodruin]

How can $z=-1$? If $\alpha=1$, then $z\in (0,2)$, no?

Exactly. The distribution should be zero there. Your expression is not only non-zero, but negative.

 

[psie]

Exactly. The distribution should be zero there. Your expression is not only non-zero, but negative.

Ok. But couldn’t we simply say that the expression for $f_Z(z)$ I gave is the distribution for $z\in (0,1+\alpha)$ and $0$ otherwise?

 

[Orodruin]

Ok. But couldn’t we simply say that the expression for $f_Z(z)$ I gave is the distribution for $z\in (0,1+\alpha)$ and $0$ otherwise?

Sure, but that is still kind of breaking it up into cases. So is using the min/max functions.