Evaluating CDF of a Random Variable with Exponential Components

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In summary, the CDF of the random variable ##X## is 1-\frac{1}{x} and the PDF is 1-\frac{1}{x}\left(1+\frac{1}{x}\right).
  • #1
EngWiPy
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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

[tex]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[/tex]

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
 
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  • #2
You can't solve this unless you know the joint distributions of the variables.
 
  • #3
micromass said:
You can't solve this unless you know the joint distributions of the variables.

All random variables are independent. I forgot to mention this.
 
  • #4
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.
 
  • #5
Is this OK?

$$
F_X(x)=Pr\left[X\leq x\right]=Pr\left[\frac{a_1}{a_2+1}\leq x\right]=\int_0^{\infty}Pr\left[a_1\leq x(\beta+1)\right]f_{a_2}(\beta)\,d\beta
$$
 
  • #6
S_David said:
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##.
 
  • #7
micromass said:
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.
 
  • #8
The integral you posted is not correct. I don't see where it comes from at all.
 
  • #9
micromass said:
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?
 
Last edited:
  • #10
I would solve it by evaluating the following integral:
[tex]\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[/tex]
 
  • #11
micromass said:
I would solve it by evaluating the following integral:
[tex]\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[/tex]

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?
 
  • #12
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?
 

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  • #13
Using the integration formula attached yields to:

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

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.
 

1. What is a CDF and why is it important to evaluate it?

A CDF (cumulative distribution function) is a function that maps the probability of a random variable taking on a value less than or equal to a given value. Evaluating a CDF allows us to determine the probabilities associated with different outcomes of a random variable, which is important in statistical analysis and decision-making.

2. How do I evaluate a CDF?

To evaluate a CDF, you need to input a specific value into the function and solve for the corresponding probability. This can be done mathematically or by using software or statistical tables.

3. What are the key properties of a CDF?

The key properties of a CDF include: it is a non-decreasing function, it ranges from 0 to 1, it is continuous, and its value at a given point represents the probability of the random variable being less than or equal to that point.

4. How can I interpret the results of evaluating a CDF?

The result of evaluating a CDF is a probability, which can be interpreted as the likelihood of a random variable taking on a value less than or equal to the input value. This can help us understand the distribution of the random variable and make predictions about its future behavior.

5. What are some common uses of evaluating CDFs?

Evaluating CDFs is commonly used in statistical analysis, risk assessment, and decision-making. It can also be used to compare different distributions and to calculate summary statistics such as mean, variance, and percentiles.

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