I have a very general question, what is probability that a low number of independent variables together are greater than some value? Suppose you are given $X_1$, $X_2$... $X_n$, where $n$ is fairly small. How can I calculate $$P(X_1+X_2+...+X_n>a)?$$ Suppose that the mean and standard deviation are known for all $X_i$. Wouldn't I still have to known the distribution of $X_1+X_2+...+X_n$? For a large $n$ the Central Limit Theorem would imply that $X_1+X_2+...+X_n$ is normal but now it can't be used, so how do I proceed?

The exact problem that I have is: $x\sim Normal(10,3)$, $Y\sim Uniform(3,7)$ and $Z\sim Exponential (20)$. What is $P(X+Y+Z>18)$?

  • $\begingroup$ In general, the distribution of the sum of independent random variables is obtained by convolution. $\endgroup$ – Zaeem Hussain Mar 7 '19 at 16:15
  • $\begingroup$ Are $X, Y, Z$ independent variables? $\endgroup$ – Peter Foreman Mar 7 '19 at 16:17
  • $\begingroup$ Yes that is also stated in the problem, sorry. $\endgroup$ – Jens Roderus Mar 7 '19 at 16:23

For small numbers you generally just have to do the sums or integrals. Let $f_X(x)$ be the probability distribution of $X$ and similarly for $Y,Z$ To have $P(X+Y+Z) \gt 18$ you do $$\int_{-\infty}^\infty f_X(x)\int_{-\infty}^\infty f_Y(x)\int_{18-X-Y}^\infty f_Z(x)dz\;dy\:dx$$ so for given values of $X,Y$ you find the chance that $Z$ is large enough. Intuitively the small block of size $dxdydz$ at $(x,y,z)$ contributes $f_X(x)f_Y(y)f_Z(z)dxdydz$ to the probability and you add up all the blocks where the sum is high enough.

  • $\begingroup$ I think you want $f_X(x)$ instead of $P(X)$ etc. $\endgroup$ – Robert Israel Mar 7 '19 at 18:11
  • $\begingroup$ @RobertIsrael: Yes, thanks $\endgroup$ – Ross Millikan Mar 8 '19 at 15:31
  • $\begingroup$ The problem is that I don't have given values of $X$ and $Y$. Now, the problem did not have to be solved analytically. So what I did was sample 1000 random values from each of the three distributions and count how many times the sums actually exceeded 18. But my question still remains. Can this be solved in theory, without assuming any values for $X$, $Y$ and $Z$? $\endgroup$ – Jens Roderus Mar 13 '19 at 8:04
  • $\begingroup$ I didn't assume values for $X$ and $Y$, I assumed known distributions. You gave an example where one was normal, one uniform, and one exponential. The integral I gave will work for any set of probability distributions. You may or may not be able to do it analytically, depending on the distributions. If you can't, a Monte Carlo approach is reasonable. For a 3D integral, 1000 points is rather few, but the idea is correct. $\endgroup$ – Ross Millikan Mar 13 '19 at 13:46

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