Let X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x) = exp(-x) , 0 <x < infinity. and f(x)= 0 elsewhere.Now, let Y1= X1+ X2.Y2 = X1/(X1+X2).Prove that Y2 and Y2 are independent. I was asked this question on an interview recently, btw.

assume X1 and X2 are independent.is the joint distribution, where you need to calculate the Jacobian: |\partial Y/\partial X| = 1/Y1. independence of Y1 and Y2 follows.

QuoteOriginally posted by: wileyswassume X1 and X2 are independent.is the joint distribution, where you need to calculate the Jacobian: |\partial Y/\partial X| = 1/Y1. independence of Y1 and Y2 follows.Yes, that is what I proved.

Try using basu's theorem. I guess it should work. No paper n front of me. Basu theorem says complete and sufficient statistic is independent of ancillary statistic

QuoteOriginally posted by: iliketopologyLet X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x) = exp(-x) , 0 <x < infinity. and f(x)= 0 elsewhere.Now, let Y1= X1+ X2.Y2 = X1/(X1+X2).Prove that Y2 and Y2 are independent. I was asked this question on an interview recently, btw.I did this problem by directly showing P{Y1<a, Y2<b}=P{Y1<a}*P{Y<b}. A somewhat lengthy method, but eventually doable.However, I would like to put this problem on its head and state it anew as follows:Let X1 and X2 be a random sample of size 2 from a distribution with probability density function f(x). Now let Y1=X1+X2 and Y2=X1/(X1+X2). Determine f(x) so that Y1 and Y2 are independent.

for the inverse question proposed above, i got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x). and when a>-1, one could easily verify there is no problem at x=0.

QuoteOriginally posted by: wileyswi got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x).The easiest way to solve is using convolution theorem and computing the Jacobian. It is almost a few lines of proof as Jacobian is just 1. Hence, f(y1,y2) = f(x1,x2) , where x1, and x2 are defined above. Since it is exponential distribution, f(y1,y2) = 1. f(y2) or something equivalent. One could use general form as well by applying the convolution theorem.

QuoteOriginally posted by: iliketopologyQuoteOriginally posted by: wileyswi got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x).The easiest way to solve is using convolution theorem and computing the Jacobian. It is almost a few lines of proof as Jacobian is just 1. Hence, f(y1,y2) = f(x1,x2) , where x1, and x2 are defined above. Since it is exponential distribution, f(y1,y2) = 1. f(y2) or something equivalent. One could use general form as well by applying the convolution theorem.In the quote above, you wrote "Since it is exponential distribution, f(y1,y2) = 1."Question: How do you know "it" is "exponential distribution"?And to follow up: What does the "it" refer to?Thank you for your posts.

QuoteOriginally posted by: quantystQuoteOriginally posted by: iliketopologyQuoteOriginally posted by: wileyswi got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x).The easiest way to solve is using convolution theorem and computing the Jacobian. It is almost a few lines of proof as Jacobian is just 1. Hence, f(y1,y2) = f(x1,x2) , where x1, and x2 are defined above. Since it is exponential distribution, f(y1,y2) = 1. f(y2) or something equivalent. One could use general form as well by applying the convolution theorem.In the quote above, you wrote "Since it is exponential distribution, f(y1,y2) = 1."Question: How do you know "it" is "exponential distribution"?And to follow up: What does the "it" refer to?Thank you for your posts."it" is r.v.Read my first post, which states what is "it" and why it is "exponential".

QuoteOriginally posted by: iliketopologyQuoteOriginally posted by: quantystQuoteOriginally posted by: iliketopologyQuoteOriginally posted by: wileyswi got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x).The easiest way to solve is using convolution theorem and computing the Jacobian. It is almost a few lines of proof as Jacobian is just 1. Hence, f(y1,y2) = f(x1,x2) , where x1, and x2 are defined above. Since it is exponential distribution, f(y1,y2) = 1. f(y2) or something equivalent. One could use general form as well by applying the convolution theorem.In the quote above, you wrote "Since it is exponential distribution, f(y1,y2) = 1."Question: How do you know "it" is "exponential distribution"?And to follow up: What does the "it" refer to?Thank you for your posts."it" is r.v.Read my first post, which states what is "it" and why it is "exponential".Your suggestion is well taken. I have read your first post -- the very first post in which the very first question is posed. It somehow seems you have not read a subsequent post by me in which a new question has been posted. In the new question there is no prior knowledge of any r.v. being exponential. When you write that "it" is r.v., you still have not explicated what "it" refers to, instead you have only qualified what kind of thing "it" is.It is very possible that I am missing something. But I am sure I would miss a lot more if the posted things are not clear to begin with.

QuoteOriginally posted by: quantystQuoteOriginally posted by: iliketopologyQuoteOriginally posted by: quantystQuoteOriginally posted by: iliketopologyQuoteOriginally posted by: wileyswi got the general form: f(x)=b^(1+a)/Gamma(1+a)*(x^a)*exp(-b*x), where a>-1 and b>0.----- ----- ----- ----- -----a sketch of my approach:assume f(x) is defined on [0, infty), so 0<=Y1<infty and 0<=Y2<=1. we also have X1=Y1*Y2 and X2=Y1*(1-Y2).since the Jacobian is Y1, the independence of Y1 and Y2 implies that f[Y1*Y2]*f[Y1*(1-Y2)] is separable, i.e., F(Y1)*G(Y2), or g[Y1*Y2]+g[Y1*(1-Y2)] can be written as F(Y1)+G(Y2), where g(x):=log[f(x)]. henceparticularly, we let Y2=1 (or 0) which yields a necessary condition: g'(Y1)-g'(0)=-Y1*g''(Y1), or d[g'(Y1)]/[g'(Y1)-g'(0)]=-dY1/Y1. integrate both sides (ignoring the singularity at Y1=0 for now), we get g(x)=a*log(x)+b*x+c, or f(x)=c*(x^a)*exp(b*x), where c (as well as the range for a and b) can be determined by normalizing f(x).The easiest way to solve is using convolution theorem and computing the Jacobian. It is almost a few lines of proof as Jacobian is just 1. Hence, f(y1,y2) = f(x1,x2) , where x1, and x2 are defined above. Since it is exponential distribution, f(y1,y2) = 1. f(y2) or something equivalent. One could use general form as well by applying the convolution theorem.In the quote above, you wrote "Since it is exponential distribution, f(y1,y2) = 1."Question: How do you know "it" is "exponential distribution"?And to follow up: What does the "it" refer to?Thank you for your posts."it" is r.v.Read my first post, which states what is "it" and why it is "exponential".Your suggestion is well taken. I have read your first post -- the very first post in which the very first question is posed. It somehow seems you have not read a subsequent post by me in which a new question has been posted. In the new question there is no prior knowledge of any r.v. being exponential. When you write that "it" is r.v., you still have not explicated what "it" refers to, instead you have only qualified what kind of thing "it" is.It is very possible that I am missing something. But I am sure I would miss a lot more if the posted things are not clear to begin with.So you are saying that that the beginning post was unclear? If so, then the original problem given to me was not any more clear. You are right that I did not read subsequent question but if you intend to change a problem somewhere in between the thread, then please start a new topic. That would make lot more sense rather than expressing frustration at people for not having read in between your lines.

QuoteOriginally posted by: iliketopologySo you are saying that that the beginning post was unclear? If so, then the original problem given to me was not any more clear.You are right that I did not read subsequent question but if you intend to change a problem somewhere in between the thread, then please start a new topic. That would make lot more sense rather than expressing frustration at people for not having read in between your lines.You make a good point. It is cleaner and definitely much less confusing to start a new thread for a new (even if closely related to a current) question. But not always does it feel natural to do so, especially when one re-poses the original question in the form of its converse. It is the nature of human communication to evolve with new ideas, new questions, unexpected changes, etc., even within the same thread. The best approach for all of us is to scan 'all' posts before we determine we are sure of what's happening. But again, with so little time at our disposal, most of us can hardly afford to do so.

QuoteOriginally posted by: quantystQuoteOriginally posted by: iliketopologySo you are saying that that the beginning post was unclear? If so, then the original problem given to me was not any more clear.You are right that I did not read subsequent question but if you intend to change a problem somewhere in between the thread, then please start a new topic. That would make lot more sense rather than expressing frustration at people for not having read in between your lines.You make a good point. It is cleaner and definitely much less confusing to start a new thread for a new (even if closely related to a current) question. But not always does it feel natural to do so, especially when one re-poses the original question in the form of its converse. It is the nature of human communication to evolve with new ideas, new questions, unexpected changes, etc., even within the same thread. The best approach for all of us is to scan 'all' posts before we determine we are sure of what's happening. But again, with so little time at our disposal, most of us can hardly afford to do so.Yep, it's a duet between the human emotion vs. the old devil - time.

i have a amateurish approach.. sadly i am an amateur actually the prob(y1<z1 and y2<z2)=p(z1&z2) can be calculated within the region bounded by x1, x2>=0, x1+x2<=z1 and x2>=(1/z2-1)x1, for the integrand exp(-x1-x2).This trivially reduces to p(z1&z2)=z2(1-exp(-z1)-z1exp(-z1)) for 0<=z2<=1 p(z1)=lt(z2 to inf-)p(z1&z2)= 1-exp(-z1)-z1exp(-z1). p(z2)=lt(z1 to inf-)p(z1&z2)=z2, hence the condition p(z1&z2)=p(z1)p(z2)for the inverse question i have the integral equation..

Since the event{X₁+X₂=y₁,X₁/(X1+X2)=y₂}is equivalent with the event{X₂=y₁(1-y&#8322,X₁=y₂y₁}we obtain for the densityf(y1,y2)=exp(-y1*(1-y2))*exp(-y1*y2) =exp(-y1)