Evaluating CDF of a Random Variable with Exponential Components

[EngWiPy]

Hello all,

I have the following random variable $X=\frac{a_1}{a_2+1}$, where $a_i=b_i/c_i$, where $b_i$ and $c_i$ are exponential random variables with mean 1. I need to evaluate the CDF of $X$ as

F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(a_2+1)\right]f_{a_2}(a_2)\,da_2

I found the CDF and PDF of $a_i$ as $F_{a_i}(x)=1-\frac{1}{1+x}$ and $f_{a_i}(x)=\frac{1}{(1+x)^2}$. My fist question is: are the limits of the integral above correct?

Thanks

 

[micromass]

You can't solve this unless you know the joint distributions of the variables.

 

[EngWiPy]

You can't solve this unless you know the joint distributions of the variables.

All random variables are independent. I forgot to mention this.

 

[micromass]

OK. I agree with the pdf and cfd of the $a_i$ then. But $a_2$ is the name of a function. So you can't use that as integration variable.

 

[EngWiPy]

Is this OK?

$$
F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(\beta+1)\right]f_{a_2}(\beta)\,d\beta
$$

 

[micromass]

Is this OK?

$$
F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(\beta+1)\right]f_{a_2}(\beta)\,d\beta
$$

I don't see how that follows. I would agree with

$$
F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(a_2+1)\right]f_{a_2}(\beta)\,d\beta
$$

But I don't see why the $a_2$ can be a $\beta$.

 

[EngWiPy]

I don't see how that follows. I would agree with

$$
F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(a_2+1)\right]f_{a_2}(\beta)\,d\beta
$$

But I don't see why the $a_2$ can be a $\beta$.

OK, thanks. So, I guess this means that the limits are correct. I would like to proceed to evaluate the integral, because I did on my papers, and got a final expression, but evaluating it using software gave me values that aren't correct.

 

[micromass]

The integral you posted is not correct. I don't see where it comes from at all.

 

[EngWiPy]

The integral you posted is not correct. I don't see where it comes from at all.

$$Pr\left[a_1\leq x(a_2+1)\right]=\int_{a_2}Pr\left[\left. a_1\leq x(\beta+1)\right| a_2=\beta\right]f_{a_2}(\beta)\,d\beta$$

To my undertsnading, since $a_2$ is a random variable too, we average over all the values of $a_2$. For the sake of argument, how would you find the CDF above?

 

[micromass]

I would solve it by evaluating the following integral:
\mathbb{P}\{a_1\leq x(a_2 + 1)\} = \int_0^{+\infty} \int_0^{x(\beta_2 + 1)} f_{a_1}(\beta_1)f_{a_2}(\beta_2) d\beta_1 d\beta_2

 

[EngWiPy]

I would solve it by evaluating the following integral:
\mathbb{P}\{a_1\leq x(a_2 + 1)\} = \int_0^{+\infty} \int_0^{x(\beta_2 + 1)} f_{a_1}(\beta_1)f_{a_2}(\beta_2) d\beta_1 d\beta_2

I think our disagreement is only about using symbols. Your evaluation can be re-written as
$$
\mathbb{P}\{a_1\leq x(a_2 + 1)\} = \int_0^{+\infty} \underbrace{\left[\int_0^{x(\beta_2 + 1)} f_{a_1}(\beta_1)\,d\beta_1\right]}_{F_{a_1}\left(x[\beta_2+1]\right)=Pr\left[a_1\leq x(\beta_2+1)\right]} f_{a_2}(\beta_2) d\beta_2
$$

which is effectively the same as I did in the the previous posts.

Are we on the same page now?

 

[EngWiPy]

Substituting the CDF of $a_1$ and the PDF of $a_2$ in the integral yields

$$F_X(x)=\int_0^{\infty}\left[1-\frac{1}{1+x(\beta+1)}\right]\frac{1}{(1+\beta)^2}\,d\beta=1-\frac{1}{x}\int_0^{\infty}\frac{1}{\beta+\left(1+\frac{1}{x}\right)}\frac{1}{(1+\beta)^2}\,d\beta$$

Then I used the integral formula attached from the table of integrals. Is there anything wrong in my work?

 

[EngWiPy]

Using the integration formula attached yields to:

F_X(x)=1-\frac{1}{x}B(1,2)_2F_1(2,1;3;1-\left(1+\frac{1}{x}\right))

where $B(.,.)$ is the Beta function, and $_2F_1$ is the Gauss Hypergeometric function. The constants in the integral attached are: $\nu=1$, $\beta=1$ which implies that $\mu=2$, and $\gamma=1+\frac{1}{x}$ which implies $\rho=1$. I checked the conditions, and they are satisfied. However, the result of $F_X(x)$ must be in $[0,\,1]$ which isn't the case when I evaluate it in Mathematica. Why? I appreciate any help.