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Let X and Y be independent random variable each having exponential distribution with parameter lambda. Then the conditional distribution X given X+Y=1 is??

How I proceeded was: X=r ,Y=1-r And accordingly I wrote the exponential distribution.

But I am doubtful about how to write the denominator. P(X+Y=1)=?

demon
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1 Answers1

1

Hint

We have $P(X+Y=1) = 0$ so you will have a hard time dividing by that. But note that we don't divide by this event when we compute the conditional density of two continuous random variables. We use the formula $$ f_{X\mid Y}(x\mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)},$$ and $f_Y(y)$ is not zero.

Method 1

You can do a standard change of variables to $$ U = X\;\;\;\;\;\; V = X + Y$$ and use the knowledge of $f_{X,Y}$ to compute the joint density $f_{U,V}(u,v).$ Then you can compute the conditional density $f_{U\mid V}(u\mid v)$ using the formula above. Setting $v=1$ will give you your desired conditional PDF for $u.$

Method 2

You can shortcut to the answer by having some knowledge and intuition about Poisson processes. $X$ and $Y$ are the waiting times between jumps. We know $X+Y=1,$ which means the second jump happened at time $1.$ So you want the conditional distribution of the time of the first jump given that the second was at time $1.$ Poisson processes have a constant jump intensity, so intuitively we'd expect that first jump to be equally likely to be anywhere in that interval $[0,1]$ that we know it occurs in. This can be proved, by method 1 or otherwise.

spaceisdarkgreen
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  • U mean the conditional distribution would follow uniform distribution?? – demon Dec 29 '17 at 08:17
  • Yes it does. Now that you know the answer it will hopefully be easier to calculate. – spaceisdarkgreen Dec 29 '17 at 08:22
  • got it.thanks for ur time – demon Dec 29 '17 at 08:24
  • I want to ask something to u.....I have come across like problems with poisson and geometric density.....They get simply solved....Y they didn't need change of variable? – demon Dec 29 '17 at 14:41
  • If you give an example of such a solution I will think about whether something like it will work here – spaceisdarkgreen Dec 29 '17 at 16:38
  • Well for e.g. X,Y ind. Poisson variate. Then to show the conditional distribution of X given X+Y is binomial. – demon Dec 30 '17 at 08:38
  • Also,let X,Y ind. Geometric variate. To show the conditional distribution of X given X+Y is uniform. Is there some discrete- continuous matter ?? I mean discrete get simply solved and continuous is not that straightforward?? – demon Dec 30 '17 at 08:40
  • The geometric is a discrete version of the exponential so the calculations are very similar. I'm guessing the 'simple' calculation you'd like to replicate is $$ P(X=k,X+Y = n) = P(X=k, Y=n-k) = p^2 (1-p)^{k-1}(1-p)^{n-k-1}=p^2(1-p)^{n-2}\ P(X+Y = n) = \sum_{k=1}^{n-1}p^2(1-p)^{n-1} = (n-1)p^2(1-p)^{n-2} \ P(X=k\mid X+Y = n) = \frac{P(X=k,X+Y = n)}{P(X+Y = n)} = \frac{1}{n-1}$$ for $1\le k \le n-1.$ – spaceisdarkgreen Dec 30 '17 at 23:12
  • The only additional complication with continuous variables is there may be a numerical factor in the joint $$f_{X,X+Y}(x,z) = f_{X,Y}(x,z-x) = \lambda^2 e^{-\lambda z}1_{x<z}.$$ Actually here, the Jacobian factor is one. Then we have $$ f_{X+Y}(z) = \int f_{X,X+Y}(x,z)dx = \lambda^2 z e^{-\lambda z}$$ and finally, $$ f_{X\mid X+Y}(x\mid z) = 1_{x<z}.$$ – spaceisdarkgreen Dec 30 '17 at 23:13
  • In fact, a lot of the work here is unnecessary. You can tell it's uniform just cause the joint distribution is constant in $x$ (or in $k$ in the first case). All the rest is just normalization. – spaceisdarkgreen Dec 30 '17 at 23:14
  • actually should be $\frac{1}{z}1_{x<z}$ for the final answer above but hopefully the process is clear. Bottom line is the way with continuous RVs isn't really any more complicated, in fact it's nearly identical to the geometric case. – spaceisdarkgreen Dec 30 '17 at 23:30