Consider independent random variables $X_1$ and $X_2$, each having density function $f_{X_{i}}(x) = 3e^{-3x}, 0\leq x<+\infty$

(i) What is the joint probability density function of ($X_1,X_2$)?

(ii) Compute the cumulative distribution function of $Z = X_1 + X_2$

(iii) Determine the probability density function of $Z$.


(i) I try to find the joint CDF before finding the joint PDF so I find $F_{X_i}(x)$ first

$F_{X_i}(x)=\int_0^x 3e^{-3t} dt=1-e^{-3x}$

But I am confused how to state the CDF of $X$:

$F_{X_i}(x) = \begin{cases} 0, & \text{if $x<0$} \\ 1-e^{-3x}, & \text{if $0\leq x < +\infty$} \\ 1, & \text{if $x\geq +\infty$} \end{cases}$

The part $x\geq +\infty$ does not even make sense

Joint CDF $F_{(X_1 , X_2)}(x_1,x_2)=F_{X_1}(x) F_{X_2}(x)$ =$\begin{cases} 0, & \text{if $x_1<0$ or $x_2 <0$} \\ (1-e^{-3x_1})(1-e^{-3x_2}), & \text{if $x_1\geq 0, x_2\geq 0$}\\ 1, & \text{if otherwise} \end{cases}$

Joint PDF would be $f_{X_1,X_2}(x_1,x_2)=\frac{\partial^2}{\partial x_1 \partial x_2}F_{(X_1 , X_2)}(x_1,x_2)$ =$9e^{x_1+x_2} (1-e^{-3x_1})(1-e^{-3x_2})$

(ii) $F_{X_1+X_2}(a)=\int_0^{+\infty} F_{X_1} (a-x_2)f_{X_2}(x_2) dx_2$

=$\int_0^{+\infty} (1-e^{-3(a-x_2)}) 3e^{-3x_2}dx_2$

Something must be wrong because the integral is unbounded.

For (i), is my answer correct? Is there better and shorter method?


1 Answer 1


Your computation of $F_{X_1,X_2}(x_1,x_2)$ is corect. By independence $f_{X_1,X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)$. No need for any differentiation.

(ii) $F_{X_1+X_2}(a)=\int_0^{a} F_{X_1} (a-x_2)f_{X_2}(x_2) dx_2$

=$\int_0^{a} (1-e^{-3(a-x_2)}) 3e^{-3x_2}dx_2$.

Note that $X_1+X_2 \leq a$ implies $X_2 \leq a$ so the integration is from $0$ to $a$. I will let you finish the computation.

(iii) Once you finish the solution of part (ii), you get the density if $Z$ by differentiation.

  • $\begingroup$ $f_{X_1,X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)=3e^{-3x_1}\cdot 3e^{-3x_2}=9e^{-3(x_1+x_2)}$ but that's not the same as the derivative. I realise there is a mistake in my derivative but even after I revise it, the answer is still different. I think $9e^{-3(x_1+x_2)}$ is the answer but I can't find the mistake in my previous working $\endgroup$
    – Magenta
    Nov 16, 2023 at 10:25
  • 1
    $\begingroup$ @Magenta $$\begin{align}\dfrac{\partial^2}{\partial x_1\,\partial x_2} \big[(1-\mathrm e^{-3x_1})(1-\mathrm e^{-3x_2})\big] &=3\mathrm e^{-3x_2}~\dfrac{\partial~~}{\partial x_1}\big[(1-\mathrm e^{-3x_1})\big]\\[1ex] &=9\mathrm e^{-3x_1-3x_2}\end{align}$$ $\endgroup$ Nov 17, 2023 at 1:50
  • $\begingroup$ Dang it, I am just dumb. Thanks mate $\endgroup$
    – Magenta
    Nov 17, 2023 at 3:40

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