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If the joint distribution $X_1$ and $X_2$ is given by $f(x_1,x_2) = 1$ for $0 < x_1 <1$ and $0 < x_2 < 1$, find the marginal distribution of the random variable $Y = X_1 + X_2$.

I am not really sure how to go about this. I know that you need to define another variable $Z = X_2$ and find the Jacobian of the two transformed variables, but I am confused on how to get the limits of the integral.

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One can use a convolution. I prefer to find the cdf of $X+1+X_2$, then differentiate. The question has been asked a number of times on MSE, but finding it is not easy. –  André Nicolas Mar 7 at 18:45

1 Answer 1

Use the transformation $$\begin{cases} Y=X_1+X_2 \\Z=X_2 \end{cases}$$ as you said. $Z$ takes values in $(0,1)$ as $X_2$ but $Y$ can take values (with positive probability) in $(0,2)$ (as the sum of $X_1$ and $X_2$). Then solve for $X_1$ and $X_2$: $$\begin{cases} X_1=Y-Z \\ X_2=Z \end{cases}$$ and calculate determinant of the Jacobian: $$\begin{vmatrix}\frac{dX_1}{dY} & \frac{dX_1}{dZ}\\ \\ \frac{dX_2}{dY}&\frac{dX_2}{dZ}\end{vmatrix}=\begin{vmatrix}1 & -1\\0& 1\end{vmatrix}=1$$

Therefore $$f_{Y,Z}(y,z)=f_{X_1,X_2}(y-z,z)\cdot|1|=1\cdot1_{0<y-z<1}\cdot1_{0<z<1}=1\cdot1_{y-1<z<y}\cdot1_{0<z<1}$$ The above condition (for the support of the random variables) can also be stated as follows: $$f_{Y,Z}(y,z)=1\cdot1_{0<y<2}\cdot1_{\max\{0,y-1\}<z<\min\{1, y\}}$$

Now, integration with respect to $z$ will give the pdf of $Y=X_1+X_2$.

  1. For $0<y<1$ we have that $$f_{X_1+X_2}(y):=f_Y(y)=\int_{0}^{y}1dz=y$$
  2. For $1\le y<2$ we have that $$f_{X_1+X_2}(y):=f_Y(y)=\int_{y-1}^{1}1dz=2-y$$

In sum $$f_{X_1+X_2}(y)=\begin{cases} y, &0<y<1\\ \\2-y, & 1\le y<2 \end{cases}$$

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