Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the difference, when calculating probabilities of Chi-Square distributions, between $<$ and $\leq$ or $>$ and $\geq$.

For example, say you are asked to find P$(\chi_{5}^{2} \leq 1.145)$.

I know that this is $=0.05$ from the table of Chi-Square distributions, but what if you were asked to find P$(\chi_{5}^{2} < 1.145)$? How would this be different?

share|improve this question
1  
There is no numerical difference since $P\{\chi_5^2 = 1.145\} = 0$ and so $$P\{\chi_5^2 < 1.145\} = P\{\chi_5^2 \leq 1.145\} - P\{\chi_5^2 = 1.145\} = P\{\chi_5^2 \leq 1.145\}.$$ Indeed, $P\{\chi_5^2 = a\} = 0$ for all real numbers $a$. –  Dilip Sarwate Feb 7 '13 at 23:56
1  
If you know calculus, you can use the fact that $$\Pr[a \le X \le b] = \int_a^b f(x) \mathrm dx$$ Consider $$\Pr[a \le X \le a] = \int_a^a f(x) \mathrm dx = 0$$ –  George V. Williams Feb 8 '13 at 0:05
add comment

1 Answer

The $\chi^2$ distributions are continuous distributions. If $X$ has continuous distribution, then $$\Pr(X\lt a)=\Pr(X\le a).$$ If $a$ is any point, then $\Pr(X=a)=0$. So in your case, the probabilities would be exactly the same.

Many other useful distributions, such as the normal, and the exponential, are continuous.

share|improve this answer
    
I'm not really following your answer. Can you elaborate on why this is so? –  Magpie Feb 8 '13 at 0:00
    
If $X$ is a random variable with density function $f(x)$, then $\Pr(X=a)=\int_a^af(x)\,dx$. The integral from $a$ to $a$ of any function is $0$. Or, in terms of areas, recall that the probability that $X$ lies between $a$ and $b$ is the area under the density function, from $a$ to $b$. Well, the area from $a$ all the way up to $a$ is $0$. –  André Nicolas Feb 8 '13 at 0:07
    
Your edit already cleared it up nicely, but that is a welcome addition too ;-) –  Magpie Feb 8 '13 at 0:08
    
Thank you for your help! –  user59633 Feb 8 '13 at 0:28
1  
You are welcome. People sometimes find this feature of continuous distributions disturbing. But one should remember that continuous distributions are not physical reality, they merely provide models, sometimes very good, sometimes not so good, of real phenomena. As an analogy, the basic formulas of Newtonian physics are a very close fit to (low velocity) reality, even though space and time may not be in fact infinitely divisible. –  André Nicolas Feb 8 '13 at 0:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.