# CDF of $\frac{X_1}{X_2X_3a}$, where $X_1$, $X_2$, and $X_3$ have exp. distributions

Problem:

Let $X_0$, $X_1$, $X_2$, and $X_3$ be four independent exponential random variables with the PDFs $f_{X_i}(x_i)={1\over \lambda_i} \exp(-\frac{x_i}{\lambda_i})$ where ($i=0,1,2,3$).

Let $X=\frac{X_0}{b}$ and $Y=\frac{X_1}{X_2X_3a}$ (where $a$, $b$ and $x$ are positive constants) and I want to find the:

\eqalign{\Pr\left(\min(X,Y)\le x\right) = \Pr(X\le x) + \Pr (Y\le x) - \Pr\left(\max(X,Y)\le x\right).\tag{1}}

The above result can be expressed in terms of the marginal distributions (only) when $X$ and $Y$ are independent, for then (1) becomes: \eqalign{\Pr\left(\min(X,Y)\le x\right) &= F_X(x) + F_Y(x) - F_X(x)F_Y(x) \\&= 1 - (1-F_X(x))(1-F_Y(x)).\tag{2}}

My attempted Soloution:

\eqalign F_X(x)=X\leq x=1-exp(-xb)

and

\eqalign {F_Y(x) &=Y\leq x \\&=\frac{X_1}{X_2X_3a}\leq x} after conditioning on $X_1$ and then on $X_3$ \eqalign {&=\int_0^\infty \frac{z_1}{X_2X_3a} \leq x \quad f_{X_1}(Z_1) \quad dz_1 \\&=\int_0^\infty\int_0^\infty X_2 \geq \frac{z_1}{z_3a} \quad f_{X_1}(Z_1) \quad f_{X_3}(Z_3) \quad dz_1 dz_3 \\ &= \int_0^\infty\int_0^\infty exp(-\frac{z_1}{z_3a\lambda_2x}) \quad f_{X_1}(Z_1) \quad f_{X_3}(Z_3) \quad dz_1 dz_3\\ &=\int_0^\infty \int_0^\infty exp(-\frac{z_1}{z_3a\lambda_2x}) {1\over \lambda_1} \exp( -\frac{z_1}{\lambda_1}) {1\over \lambda_3} \exp(-\frac{z_3}{\lambda_3}) dz_1 dz_3 \\}

Since ${1\over \lambda_1}\int_0^\infty exp(-\frac{z_1}{z_3a\lambda_2x}) \exp( -\frac{z_1}{\lambda_1}) dz_1 = \frac{z_3}{z_3+(a\lambda_1 x)}$,

let ${1\over \lambda_3}=k$ and $a\lambda_1 x=c$ then, \eqalign {&={k}\int_0^\infty exp(-{z_3k}) \frac{z_3}{z_3+(c)} dz_3 \\ &=-kc \exp(ck)E_1(ck)+1} where $E_1(ck)$ is the exponential integral function.

Therefore, the $min(X,Y)\leq x = 1 -\Big(exp(-xb) (kc \exp(ck)E_1(ck)\Big)$.

Any comments or pointing out mistakes will be very much appropriated.

• I am afraid you have some notational difficulties. For example, what does your equation 2 mean? What are you trying to find, the distribution of $\min (X,Y)$? What is $x$, and how can it equal $1 - (1-F_X(x))(1-F_Y(x))$? (It can, but possibly only at $x=0$.) I think a significant rewrite will be needed to make sense of what you're asking. – jbowman Oct 7 '17 at 16:17
• You might find the work easier if you were to express the event $Y\gt x$ as $X_1\gt X_2X_3ax$: there would be no quotients to work with. Evidently the exponential integral would not arise in computing the chance of the latter event, suggesting something went wrong in your calculations. – whuber Oct 7 '17 at 16:37
• @jbowan, Thank you for your kind reply. I have updated the question description. Can you kindly check it out again? – Baidal Kocham Oct 8 '17 at 4:23
• @whuber, Thank You sir. I have now updated the question. Can you please look at it again? – Baidal Kocham Oct 8 '17 at 4:26