Exponential Distribution. Let x be a random variable from an exponential distrib

Question

Exponential Distribution. Let x be a random variable from an exponential distribution, i.e fx(x) ce-x/2u(x) = where c is a constant and u(r) is a step function. 1. Find c and sketch fr(x) 2. Find and sketch Fx(a) 3. Find F.(ax21) 4. Find fa(al2 1) 5. Find F, (a|x = 1) and f. (alx = 1) 6. Find E x^ 7. Find E a2) 8. Find 9. Find cov(x,y) for y = 2x. Find the corsponding correlation coefficient. 10. This type of distribution is often used to model the amount of time between consecutive random events, e.g., email messages arriving at a mail server, customers arriving at a counter, etc. In this case, x represents the time variable. For the distribution above, (assume x is in seconds) what is the average number of arrivals per second? (hint: you already know the average time between arrivals)

Explanation / Answer

a) c can be found out by integrating the probability and setting it equal to 1. f(x)=c*e^(-x/2)u(x). u(x) is 1 when x>0 and 0 other wise. Thus integrating for x(0, infinity) we get f(x)dx= -2c*e^(-x/2) and thus putting the limits we get 2c=1 or c=1/2.

b)F(x,a) is the cumulative distribution of f(x)dx integrating from (0,a), thus substituting the value of c and integrating we get f(x)dx= 1/2*e^(-x/2)dx or -e^(-x/2) and then putting limits we get F(a)= 1-e^(-a/2). for a>0 and 0 for a<0.

c) given x>=1, we can integrate the above function f(x) from (1,a) and thus F(a|x>=1)=e^(-1/2)-e^(-a/2) for a>1 and for a<1, it is 0.

d)f(x) for x>=1 is simply 1/2*e^(-a/2) for a>1 and 0 otherwise.

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