# (Solution-verification) Transformation of Joint Probability 3 independent variables case

Let X_1, X_2, X_3 are independent random variables and let the probability density of each variable be as following

$$f(x_i) = e^{-x}I_{(0,\infty)}(x)$$

Now, First I would like to derive joint probability density function of $Y_1 = X_1, Y_2 = X_1+X_2, Y_3 = X_1+X_2+X_3$

my solution

the domain of each $Y_i$ is $(0,\infty)$ by the definition of each random variable and the domain of each $X_i$

Then since joint probability of $X_1, X_2, X_3$ is $g(x_1,x_2,x_3) = e^{-x_1-x_2-x_3}I_{(0<x_i<\infty, \;i = 1,2,3)}$ by independence between $X_i$

Then if $X_i$ are substituted with random variables $Y_i$ by $X_1 = Y_1, X_2 = Y_2-Y_1, X_3 = Y_3 - Y_1- Y_2$, $h(y_1,y_2,y_3) = e^{y_1-y_3}I_{(0<y_1, y_3\infty)}$

Second, I need to derive joint pdf of $Y_1$ and $Y_3$

Since $X_3 = Y_3-Y_1-Y_2$ One can derive $f(y_3-y_1-y_2) = e^{-y_3+y_1-y_2}I_{(0<y_i, \;i=1,2,3<\infty)}$ then the marginal probability about y_2 would be $\int_0^\infty e^{-y_3+y_1-y_2}I_{(0<y_i, \;i=1,2,3<\infty)}dy_2=e^{y_1-y_3}I_{(0<y_1, y_3<\infty)}$

From the first and second, it is revealed that joint pdf $Y_1, Y_2, Y_3$ equals to its joint pdf of $Y_1$ and $Y_3$

(I don't know what it means more mathematically anyway, in other words any specific term or expression we refer to upon this characteristic?)

• $X_3 = Y_3-Y_2$, isn't it? – Łukasz Deryło Jul 14 '17 at 5:50

Use moment generating function will be easier for this problem.

Since $f(x) = e^{-x}I_{(0,\infty)}(x)$

$$M(t)=E(e^{tX})=\int_0^{\infty}e^{tx}e^{-x}dx=\frac{1}{1-t}, (t<1)$$

For $Y_1=X_1$ there are not a lot to say, the pdf is just $f(y) = e^{-y}I_{(0,\infty)}(y)$, just change the symbol of the variable.

For $Y_2=X_1+X_2$

$M(t)=E(e^{tY_2})=E[e^{t(X_1+X_2)}]=E(e^{tX_1})*E(e^{tX_2})=\frac{1}{(1-t)^2}$ by independent.

So $Y_2$ has a gamma distribution with $\theta=1$ and $\alpha=2$.

I think now you can do the same with $Y_3=X_1+X_2+X_3$

• Third case, the answer would be $Y_3 \thicksim Gamma(3,1)$ from its mgf $1/(1-t)^3$ – Daschin Jul 16 '17 at 9:10
• But in your answer, it only covers pdf of each $Y_i$. How could one derive joint pdf of$Y_i$s? – Daschin Jul 16 '17 at 9:16
• OK, I did not read your question carefully. If you want joint pdf of $Y_1, Y_2, Y_3$, I think you better use Jacobin transformation. I will see if I can help later. – Deep North Jul 17 '17 at 0:59
• Thx acutally your approach way simplify my task for other problem sets. – Daschin Jul 17 '17 at 1:52