Given x is an exponential random variable, find median & probability

Question

For the median, I believe that I should integrate the function, ∫x0λe−λtdt=1−e−λx Then I need 1−e−λm=.5 for m, which is equivalent to e−λm=.5. m=ln(2)/λ =>m=ln(2)/.2

Answer 1

a) For the median, I believe your computation is correct, my only doubt is because of the lack of TeX. We want $m$ such that $$\int_0^m \lambda e^{-\lambda x}\,dx=\frac{1}{2}.$$ Integrate. We want $$1-e^{-\lambda m}=\frac{1}{2},$$ or equivalently $$e^{-\lambda m}=\frac{1}{2}.$$ Take the logarithm of both sides. We get $$-m\lambda=\ln(1/2)=-\ln 2,$$ and therefore $m=\frac{\ln 2}{\lambda}$.

b) The probability that $X_1=X_2$ is $0$. (It is the integral over the line $x_1=x_2$ of the joint density function.)

By symmetry, $\Pr(X_1\gt X_2)=\Pr(X_2\gt X_1)$. By symmetry each is equal to $\frac{1}{2}$.

Or else we could integrate. The joint density function of $X_1$ and $X_2$ is $\lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}$ for $x_1\gt 0$, $x_2\gt 0$, and $0$ elsewhere. Thus $$\Pr(X_2\gt X_1)=\int_{x_1=0}^\infty \left(\int_{x_2=x_1}^\infty \lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}\,dx_2 \right)\,dx_1.$$ After some calculation we get $\frac{1}{2}$. Exploiting symmetry is a lot easier!

c) The probability that $Y\gt 3$ is equal to the probability both $X_1$ and $X_2$ are $\gt 3$. The probability that $X_1$ is greater than $3$ is $\int_3^\infty \lambda e^{-\lambda x_1}\,dx_1$. This is $e^{-3\lambda}$. For the probability that $Y\gt 3$, square.

d) I do not know what tools you are expected to use, possibly the memorylessness of the exponential. Then we want the probability that an exponential with parameter $\lambda$ is $\gt 8$ given that it is $\ge 3$. This is the probability that an exponential with parameter $\lambda$ is $\gt 5$.