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I have two random variables X and Y which came from different inverse gaussian (IG) distributions: $$ X \sim IG(\mu_1,\lambda_1)$$ $$ Y \sim IG(\mu_2,\lambda_2)$$

I need to find a probability of X>Y. It seems, that i need to find a pdf and cdf for variable Z=X-Y

$$ p(X-Y|\mu_1,\mu_2,\lambda_1,\lambda_2)$$

Is it possible to find and analytical solution for this problem? I will need these functions for further MLE estimation.

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    $\begingroup$ You're likely to make more progress by evaluating $\Pr(X/Y \gt t)$ for general $t\gt 0,$ because that immediately eliminates both the shape parameters. $\endgroup$
    – whuber ♦
    May 15, 2022 at 15:55
  • $\begingroup$ @whuber I'm clearly missing something. I get $\frac{e^{\frac{\lambda_1}{\mu_1}+\frac{\lambda_2}{\mu_2}} \sqrt{\frac{\lambda_1 \lambda_2 \left(\lambda_2 \mu_1^2+\lambda_1 \mu_2^2 t\right)}{\lambda_1+\lambda_2 t}} K_1\left(\frac{\sqrt{\frac{(\lambda_1+t \lambda_2) \left(\lambda_2 \mu_1^2+t \lambda_1 \mu_2^2\right)}{t}}}{\mu_1 \mu_2}\right)}{\pi \mu_1 \mu_2 t}$ for the pdf of $T=X/Y$ so I don't see finding 1 minus the cdf for $t=1$ eliminates the both of the shape parameters. $\endgroup$
    – JimB
    May 15, 2022 at 20:00
  • $\begingroup$ And $K_1$ is the modified Bessel function of the second kind: $K_n(z)$ with $n=1$. $\endgroup$
    – JimB
    May 15, 2022 at 20:09
  • $\begingroup$ @whuber Thank you for a nice idea. Unfortunately, my knowledge of algebra is rather poor and I couldn't figure out how I could derive a pdf and cdf for this value. I tried to use these links for this: en.wikipedia.org/wiki/Ratio_distribution; mathworld.wolfram.com/RatioDistribution.html; math.stackexchange.com/questions/4292320/… $\endgroup$
    – zlon
    May 16, 2022 at 6:24
  • $\begingroup$ @JimB Generally, suppose $X/\lambda_x$ is assumed to be in one distributional family and $Y/\lambda_y$ in another family (which could be the same) and both families are supported on the positive reals. Because the event $X\gt Y$ is equivalent to $X/Y\gt \lambda_x/\lambda_y=t,$ the question of finding the chance of this event is reduced to that of finding the CDF of $X/Y.$ $\endgroup$
    – whuber ♦
    May 16, 2022 at 12:57

1 Answer 1

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As mentioned by @whuber, you're better off attempting to obtain the distribution of $X/Y$ than $X - Y$. Because both $X$ and $Y$ are supported on the positive reals one can write $Pr(X>Y)$ as $Pr(X/Y > 1)=$. So this will involve obtaining the cdf of $X/Y$.

Using Mathematica (and Maple and MATLAB and others can certainly do the same) I get the following for the pdf:

dist = TransformedDistribution[x/y, 
 {x \[Distributed] InverseGaussianDistribution[n1, s1], 
  y \[Distributed] InverseGaussianDistribution[m2, s2]}];
pdf = Simplify[PDF[dist, t], Assumptions -> t > 0]

pdf of X/Y

I don't believe that there's a nice closed form for the cdf so you'll need use numerical integration to get the cdf. Assuming you might be doing this in R:

# Define pdf
  pdf <- function(t, m1, s1, m2, s2) {
    (exp(s1/m1 + s2/m2)*
     sqrt((s1*s2*(m1^2*s2 + m2^2*s1*t))/(s1 + s2*t))*
     besselK(sqrt(((m1^2*s2 + m2^2*s1*t)*(s1 + s2*t))/
     t)/(m1*m2),1))/(m1*m2*pi*t)            
  }
  
# Define cdf
  cdf <- function(t0, m1, s1, m2, s2) {
    1 - integrate(pdf, 0, t0, m1=m1, s1=s1, m2=m2, s2=s2)$value  
  }
  
# Pr(X > Y)
  1 - cdf(1, 1, 2, 2, 3)
# [1] 0.7420883

A partial simplification

We have $X \sim IG(\mu_1, \lambda_1)$ and $Y\sim IG(\mu_2,\lambda_2)$ and want to determine $Pr(X>Y)$. Note that we can write $X=\lambda_1 X’$ where $X’\sim IG(\mu_1/\lambda_1, 1)$ and $Y=\lambda_2 Y’$ where $Y’\sim IG(\mu_2/\lambda_2,1)$. We have

$$Pr(X>Y)=Pr(\lambda_1 X’ > \lambda_2 Y’)=Pr\left(\frac{X’}{Y’}>\frac{\lambda_2}{\lambda_1}\right)$$

If we let $\rho_1=\frac{\mu_1}{\lambda_1}$ and $\rho_2=\frac{\mu_2}{\lambda_2}$, then the pdf for $X’/Y’$ is given by

$$\frac{e^{\frac{1}{\rho_1}+\frac{1}{\rho_2}} \sqrt{\frac{\rho_1^2+\rho_2^2 t}{t+1}} K_1\left(\frac{\sqrt{\frac{(t+1) \left(\rho_1^2+t \rho_2^2\right)}{t}}}{\rho_1 \rho_2}\right)}{\pi \rho_1 \rho_2 t}$$

There doesn’t appear to be a closed form for the cdf so numerical integration will need to be used. One would need to evaluate 1 minus the cdf evaluated at $\lambda_2/\lambda_1$ to find $Pr(X>Y)$. So with this simplification to achieve the objective (finding $Pr(X>Y)$), we only need 3 parameters rather than 4: $\mu_1/\lambda_1$, $\mu_2/\lambda_2$, and $\lambda_2/\lambda_1$. However, given the likely need to estimate all 4 parameters ($\mu_1$, $\lambda_1$, $\mu_2$, and $\lambda_2$) there isn’t much of a savings.

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  • $\begingroup$ The whole point to the scale parameter observation is that it simplifies the problem, eliminating two of the four parameters. Unfortunately, the difficulty with integrating the pdf persists. $\endgroup$
    – whuber ♦
    May 16, 2022 at 17:33
  • $\begingroup$ @JimB Thank you very much for your answer! Unfortunately, now I'm not so sure I understood and used it correctly. Could you look up the "successor" question to this one. stats.stackexchange.com/questions/575591/… $\endgroup$
    – zlon
    May 17, 2022 at 11:56

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