# Distribution of $X+Y$ when $X$ and $Y$ are i.i.d with pdf $f(x)=\alpha\beta^{-\alpha}x^{\alpha-1}\mathbf1_{0<x<\beta}$

I am working on the following problem:

Let $X$ and $Y$ be independent random variables with common density $f(x)=\alpha\beta^{-\alpha}x^{\alpha-1}\mathbf1_{0<x<\beta}$ where $\alpha\geqslant1$. Let $U=\min(X,Y)$ and $V=\max(X,Y)$. Find the joint density of $(U,V)$ and hence find the pdf of $U+V$.

As $U+V=X+Y$, I can simply find the pdf of $X+Y$ to see what the pdf of $U+V$ should be.

I get the pdf of $T=X+Y$ to be $$f_T(t)=\int f(t-y)f(y)\,\mathrm{d}y=\alpha^2\beta^{-2\alpha}\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y\,\mathbf1_{0<t<2\beta}\tag{1}$$

Not sure if that integral can be simplified though.

Coming back to the actual question, the joint pdf of $(U,V)$ is given by

$$f_{U,V}(u,v)=2f(u)f(v)\mathbf1_{0<u<v<\beta}=2\alpha^2\beta^{-2\alpha}(uv)^{\alpha-1}\mathbf1_{0<u<v<\beta}$$

I did a change of variables $(U,V)\to(W,Z)$ where $W=U+V$ and $Z=U$. Absolute value of jacobian is unity. Also, $0<u<v<\beta\implies 0<z<\frac{w}{2}<\beta$. So marginal pdf of $W$ is

$$f_W(w)=2\alpha^2\beta^{-2\alpha}\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}\,\mathbf1_{0<w<2\beta}\tag{2}$$

It is possible that I have made some error in the proper supports of the random variables. It is also possible that the integral does not have a solution in terms of elementary functions. In any case, I could not proceed with the integral. So I couldn't even verify that $W=U+V$ has the same pdf as $T=X+Y$. It appears that I am getting different distributions of $W$ and $T$. And out of curiosity, does the distribution of $X$ have a name (in which case I would have searched for the convolution of two such random variables) ?

Edit.

Proceeding with the last integral I get by hand

$$\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}=w^{2\alpha-1}\int_0^{1/2}t^{\alpha-1}(1-t)^{\alpha-1}\,\mathrm{dt}=w^{2\alpha-1}I_{1/2}(\alpha,\alpha)B(\alpha,\alpha)$$ where $I_{x}$ is the regularized incomplete beta function. Using the property $I_x(a,b)=1-I_{1-x}(b,a)$, we get $I_{1/2}(\alpha,\alpha)=\frac{1}{2}$. So finally we have $$\int_0^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{dz}=\frac{1}{2}w^{2\alpha-1}B(\alpha,\alpha)$$

This implies that

$$f_W(w)=\alpha^2\beta^{-2\alpha}B(\alpha,\alpha)w^{2\alpha-1}\mathbf1_{0<w<2\beta}$$

That this is not a density in the given range of $w$ is easily seen. So I feel I have made a big mistake somewhere. I have checked my calculations with Mathematica and they seem to agree.

• @Xi'an And the sum of independent beta variates does not have a closed form pdf perhaps? Apr 14, 2018 at 13:03
• @Xi'an So I feel there's nothing wrong if I end my answer with that integral irrespective of whether it has a closed form in terms of some special function or not? Apr 14, 2018 at 13:09
• As a generalization of stats.stackexchange.com/questions/41467 (the case where $\alpha=1$), this question likely can be solved using one or more of the various techniques explained in that thread.
– whuber
Apr 14, 2018 at 17:41
• I mistakenly said that $\alpha>1$, when in fact $\alpha>0$ suffices for $f$ to be a valid density. This is sometimes called a power function distribution. For $\beta=1$ it is a beta density, and for $\alpha=1$ it is a uniform density. Nov 8, 2018 at 11:29

Since \begin{align} \int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y\,\mathbf1_{0<t<2\beta}&=\begin{cases} \int_{0}^{t}(y(t-y))^{\alpha-1}\,\mathrm{d}y&\text{when }0\le t\le \beta\\ \int_{t-\beta}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y&\text{when }\beta\le t\le 2\beta\\ \end{cases}\\ \end{align} we have ($t<\beta$) $$\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y= \int_{0}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y+\int_{t/2}^{t}(y(t-y))^{\alpha-1}\,\mathrm{d}y$$ and by a change of variable $z=t-y$ in the second integral of the rhs $$\int_{\max(t-\beta,0)}^{\min(t,\beta)}(y(t-y))^{\alpha-1}\,\mathrm{d}y= 2\int_{0}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y$$ Similarly, when $t>\beta$, \begin{align*} \int_{t-\beta}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y&= \int_{t-\beta}^{t/2}(y(t-y))^{\alpha-1}\,\mathrm{d}y+\int_{t/2}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y\\ &=2\int_{t/2}^{\beta}(y(t-y))^{\alpha-1}\,\mathrm{d}y \end{align*} again by a change of variable $z=t-y$ in the second integral of the rhs. I am however unable to recover the same functional expression for the density in this second case, namely$$2\int_{0}^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{d}z$$
Now, as pointed out in the question, $$2\int_{0}^{w/2}(z(w-z))^{\alpha-1}\,\mathrm{d}z \propto w^{2(\alpha-1)+1}=w^{2\alpha-1}$$by a change of scale, which would imply that the distribution of interest has the density $$f(w)\propto w^{2\alpha-1} \mathbf1_{0<w<2\beta}$$ which turns it into a Beta ${\cal B}(2\alpha,1)$ distribution rescaled on $(0,2\beta)$, hence with density $$f(w) = \{2\beta\}^{-2\alpha}\dfrac{\Gamma(2\alpha+1)}{\Gamma(2\alpha)}w^{2\alpha-1} \mathbf1_{0<w<2\beta}=2\alpha\{2\beta\}^{-2\alpha}w^{2\alpha-1} \mathbf1_{0<w<2\beta}$$
This comes as a contradiction when considering the unbelievably detailed answer from W. Huber, since Uniforms are Beta ${\cal B}(1,1)$. And since the sum of two Uniforms is not a Beta ${\cal B}(2,1)$ random variable, but instead an rv with "tent" density.
The issue is thus with the derivation of the density of $W=U+V$: since $$(U,V) \sim 2\alpha \beta^{-2}[uv]^{\alpha-1}\,\mathbb{I}_{0<u<v<\beta}$$ a change of variables $(Z,W)=(U,U+V)$ leads to $$(Z,W) \sim 2\alpha \beta^{-2}[z(w-z)]^{\alpha-1}\,\mathbb{I}_{0<z<w-z<\beta}$$ and the indicator constraints are $$0<z \quad 2z<w \quad z<\beta \quad z>w-\beta \quad 0<w \quad\text{and}\quad w<2\beta$$ Therefore, in conclusion, $$W\sim 2\alpha^2 \beta^{-2\alpha}\int_{\max\{0,w-\beta\}}^{\min\{\beta,w/2\}}[z(w-z)]^{\alpha-1}\,\text{d}z\,\mathbb{I}_{0<w<2\beta}$$ namely (1) and not the proposed expression (2).
• That was what I was asking whether it agrees with (1) or not. You probably have to add the missing constants in $(Z,W)$ and $(U,V)$ as well. Thanks, no wonder I was getting all those weird results. Apr 16, 2018 at 13:54