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.

Let $X$ be an exponential random variable with parameter $\lambda > 0$. Find the probabilities $$P(X>2/\lambda)$$ and $$P(|X-1/\lambda|<2/\lambda)$$

I am not sure where to start with this. I know that $P(X>x)=1-F(x)=e^{-\lambda x}$. Does this help with the first one in some way? Do I plug $2/\lambda$ in for $x$ and that gets rid of the $\lambda$'s so the answer would be ${e^-}^2$? Lost on the second.

share|improve this question
    
Is this homework? –  S4M Sep 26 '12 at 12:43
    
@S4M-Is is. Why? –  Sprock Sep 26 '12 at 16:21
1  
Well normally there is a tag for homeworks. When questions are for homework people often want to just give a clue to the solution, instead of doing the whole work for the OP. –  S4M Sep 26 '12 at 19:34
    
In the future I will include the tag. Didnt know sorry. –  Sprock Sep 26 '12 at 21:46
add comment

1 Answer

up vote 5 down vote accepted

Yes, the fact that you know that $$\Pr(X\gt x)=e^{-\lambda x}$$ (if $x\ge 0$) is very useful. Equivalently, $$F(x)=1-e^{-\lambda x}$$ when $x\ge 0$, and $F(x)=0$ elsewhere. Being aware of the "elsewhere" part can help avoid error.

For the first problem, just substitute $2/\lambda$ for $x$. Note that $\lambda(2/\lambda)=2$.

For the second, let's "unwrap" what $|X-1/\lambda|\lt 2/\lambda$ says. It says that $X$ is not too far from $1/\lambda$, indeed within $2/\lambda$ of $1/\lambda$. More precisely, it says that $$\cfrac{1}{\lambda}-\cfrac{2}{\lambda} \lt X \lt \cfrac{1}{\lambda}+\cfrac{2}{\lambda}.$$ Some simplification reduces this to $$-\cfrac{1}{\lambda}\lt X\lt \cfrac{3}{\lambda}.$$ But note that an exponentially distributed random variable can never be negative. So all we want is $$\Pr\left(X\lt \cfrac{3}{\lambda}\right).$$ Another way of arriving at the answer is to note that our probability is $F(3/\lambda)-F(-1/\lambda)$. But now we must be careful. Since $-1/\lambda$ is negative, $F(-1/\lambda)=0$, so our answer is just $F(3/\lambda)$. We get this by substituting $3/\lambda$ for $x$ in the formula for $F(x)$.

share|improve this answer
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.