Question

Thus, the expected time waiting is 5/6 hours (or 50 minutes) (Note that it is wrong to reason like this: Alice expects to arrive at 12:30, Bob expects to arrive at 1:00: thus, we expeet that Bob will wait 30 mimutes for Alice.) (b). We want to compute the probability that Bob has to wait for Alice, which is P(Y < X), which we do by integrating the joint density, f(r,y), over the region where < x. Draw a picture! (Show the support set (a rectangle), and the line y = z.) Erample 12: Suppose X and Y are mdependent and that X is exponential with mean 0.5 and Y has density 0 otherwise Find the density of the rundom variable W - miniXY) and the random variable z - X+Y CFO POE 12ge.pp..pp.. CFO POE 12ge.pp

Can you solve 12

Answer 1

$CDF~of~W~is\\ F_W(w)=P(\min\{X,Y\}\leq w)=1-P(\min\{X,Y\}>w)\\ =1-P(X>w)P(Y>w),~since~X~and~Y~are~independent\\ =1-\left(\int_w^\infty 2\exp(-2x)dx \right )\left(\int_w^\infty 3\exp(-3y)dy \right )\\ =1-\exp(-2w)\exp(-3w)=1-\exp(-5w),~w>0\\ PDF~of~W~is\\ f_W(w)=\frac{\mathrm{d} F_W(w)}{\mathrm{d} w}=5\exp(-5w),~w>0\\ =0~otherwise\\ Hence~W\sim Exponential~distribution~with~parameter~5.\\ The~pdf~of~Z~is\\ f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)dx\\ where,~f_X(x)=2\exp(-2x),~x>0\\ =0~otherwise\\ and~f_Y(z-x)=3\exp(-3(z-x)),~z-x>0~or,~z>x>0\\ =0~otherwise\\ Hence~f_Z(z)=6\exp(-3z)\int_0^z\exp(x)dx\\ =6\exp(-3z)\left(\exp(z)-1 \right ),~z>0\\ =0~otherwise$