# Finding Joint PDF, CDF of $X_1 + X_2$ and PDF of $X_1 + X_2$ from density function of X

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 $$Z$$.

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(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

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.

• $f_{X_1,X_2}(x_1,x_2)=f_{X_1}(x_1)f_{X_2}(x_2)=3e^{-3x_1}\cdot 3e^{-3x_2}=9e^{-3(x_1+x_2)}$ 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 Nov 16, 2023 at 10:25
• @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} Nov 17, 2023 at 1:50
• Dang it, I am just dumb. Thanks mate Nov 17, 2023 at 3:40