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Let random variables $X_1$, $X_2$ and $X_3$ be independent and identically distributed according to the exponential distribution with rate $\lambda$. Let $Y_1 = X_1$, $Y_2 = X_1 + X_2$, and $Y_3 = X_1 + X_2 + X_3$.

(a) Find the joint density function of $Y_1$, $Y_2$ and $Y_3$.

(b) Find the marginal density of $Y_3$.

What do I do here? I don't really know where to start. Thank you for any help!

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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Kavi Rama Murthy May 04 '21 at 09:24
  • I know about joint pdf pmfs etc. but not a "joint density" or a "marginal density". I also don't see how you can add random variables? If an rv A was tossing a coin and B was rolling a die how does let C = A+B make sense? You can't add coin tosses and die rolls! – Adam Mac. May 04 '21 at 09:30
  • If you roll two dice, you can add to the values together (for games like craps, that is precisely what you do). – Henry May 04 '21 at 09:32
  • Yes, but this is exponential distribution so is Y3 just 3(λe^(-λx))? – Adam Mac. May 04 '21 at 09:35

2 Answers2

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(b) If $X_1, \dots, X_n \stackrel{iid}{\sim}\mathsf{Exp}(\mathrm{rate} = \lambda),$ then $T = \sum_{i=1}^n X_i \sim\mathsf{Gamma}(\mathrm{shape}= n, \lambda).$ This is easily proved using moment generating functions. (See Wikipedia on gamma distributions; the second parameterization, using rate, is used in R.)

By simulation of $m = 100\,000$ realizations of $T,$ each from summing $n=5$ independent $X_i \sim \mathsf{Exp}(.25),$ we have the following demonstration that $T\sim\mathsf{Gamma}(5, .25)$ with $E(T) = n/\lambda = 20.$

set.seed(504)
t = replicate(10^5, sum(rexp(5, .25)))
summary(t)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.6922 13.5211 18.7286 20.0232 25.0858 81.6024

hdr="Simulated sample of GAMMA(5, .25)" hist(t, prob=T, br=20, col="skyblue2", main=hdr) curve(dgamma(x, 5, .25), add=T, col="orange", lwd=2)

enter image description here

BruceET
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Joint density means a specification of the density of the random variables as they would appear together, i.e. $g(Y_1, Y_2, Y_3)$. Marginal density means the density of one such variable alone, $g(Y_3)$.

a) Let us use the standard transformation lemma.

Inverse mappings give

$$\begin{split}s_1(Y_1, Y_2, Y_3)&=X_1=Y_1\\ s_2(Y_1, Y_2, Y_3)&=X_2=Y_2-Y_1\\ s_3(Y_1, Y_2, Y_3)&=X_3=Y_3-Y_1-(Y_2-Y_1)=Y_3-Y_2\end{split}$$

The absolute value of the Jacobian is, expanding across the first row, $$|J(s_1(Y_1,Y_2,Y_3), s_2(Y_1,Y_2,Y_3), s_3(Y_1,Y_2,Y_3))|=\operatorname{abs}\left(\begin{vmatrix}1&0&0\\-1&1&0\\0&-1&1\end{vmatrix}\right)=\operatorname{abs}\left(1\begin{vmatrix}1&0\\-1&1\end{vmatrix}\right)=|1|=1$$

The joint density of $X_1, X_2, X_3$ is

$$f(X_1,X_2,X_3)=\lambda e^{-\lambda X_1}\lambda e^{-\lambda X_2}\lambda e^{-\lambda X_3}$$

So the joint density of $Y_1, Y_2, Y_3$ is given by plugging in the inverse mappings and multiplying by the absolute value of the Jacobian,

$$\begin{split}g(Y_1,Y_2,Y_3)&=f(s_1(Y_1,Y_2,Y_3), s_2(Y_1,Y_2,Y_3), s_3(Y_1,Y_2,Y_3))\cdot |J(s_1(Y_1,Y_2,Y_3),s_2(Y_1,Y_2,Y_3),s_3(Y_1,Y_2,Y_3))|\\ &=\lambda^3e^{-\lambda(Y_1+Y_2-Y_1+Y_3-Y_2)}\cdot 1\\ &=\lambda^3e^{-\lambda Y_3}\end{split}$$

For what values are valid, we find that exponential rv's are only positive so

$$Y_1>0,Y_2-Y_1>0,Y_3-Y_2>0\Rightarrow Y_3>Y_2>Y_1>0$$

b) As mentioned (+1), the sum of $n$ independent exponential random variables with the same rate is $\text{Gamma}(n, \lambda)$, so here it is $Y_3\sim\text{Gamma}(3, \lambda)$. The density is $g(y_3)=\frac{\lambda^3}{2}x^2e^{-\lambda y_3},y_3>0$.

Vons
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  • Nice presentation. – BruceET May 04 '21 at 19:40
  • Yeah thank you, I don't see where the matrix came from, or know what an inverse mapping is but thank you. Where'd the Gamma come from? – Adam Mac. May 05 '21 at 07:35
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    It’s the name of a type of distribution – Vons May 05 '21 at 13:15
  • Yes but we are dealing with the exponential distribution not the gamma distribution. – Adam Mac. May 05 '21 at 18:19
  • exponential is a special case of gamma being $\text{Exponential}(\lambda)\equiv\text{Gamma}(1, \lambda)$. It is known that when you add exponential distributions with the same parameter, it turns into gamma and can no longer be represented as exponential. – Vons May 05 '21 at 18:29