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carlo
carlo
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This is actually a bayesian problem. Time $Y_1$ if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of $Y_1$. Applying Bayes rule:

$$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$

So you have the exponential distribution $p(Y_1)$ that we already understand, and the other stranger thing $p(emails_1 = 1|Y_1)$ which is the probability of not having any other mail sent after the first one, until time 1, given $Y_1$. It is equal to $\int_1^\infty p(Y_2|Y_1) dY_2$ and it raises as $Y_1$ gets closer to 1. Actually, as the integral of an exponential function is exponential as well, it turns out, with few passages, that $p(emails_1 = 1|Y_1) = p(1-Y_1)$$p(emails_1 = 1|Y_1) = exp(1-Y_1)$, so:

$$p(Y_1|emails_1 = 1) \propto exp(Y_1)exp(1-Y_1) \propto 1$$

where $exp$ is the esponential distribution density, whatever is the (equal) rate parameter. Last passage is easily verifiable from multiplying those exponential densities.

Your updated distribution of $Y_1$ is uniform, and expected value can be immediately derived.

If you'd like a much shorter, more intuitive explanation, mind that Poisson process is symmetrical, and that the distribution of $Y_1$ is the same looked from 0 and from 1, so of course its expected value must be 0.5. This explanation doesn't account for the rest of its distribution though.

This is actually a bayesian problem. Time $Y_1$ if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of $Y_1$. Applying Bayes rule:

$$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$

So you have the exponential distribution $p(Y_1)$ that we already understand, and the other stranger thing $p(emails_1 = 1|Y_1)$ which is the probability of not having any other mail sent after the first one, until time 1, given $Y_1$. It is equal to $\int_1^\infty p(Y_2|Y_1) dY_2$ and it raises as $Y_1$ gets closer to 1. Actually, as the integral of an exponential function is exponential as well, it turns out, with few passages, that $p(emails_1 = 1|Y_1) = p(1-Y_1)$ so:

$$p(Y_1|emails_1 = 1) \propto exp(Y_1)exp(1-Y_1) \propto 1$$

where $exp$ is the esponential distribution density, whatever is the (equal) rate parameter. Last passage is easily verifiable from multiplying those exponential densities.

Your updated distribution of $Y_1$ is uniform, and expected value can be immediately derived.

If you'd like a much shorter, more intuitive explanation, mind that Poisson process is symmetrical, and that the distribution of $Y_1$ is the same looked from 0 and from 1, so of course its expected value must be 0.5. This explanation doesn't account for the rest of its distribution though.

This is actually a bayesian problem. Time $Y_1$ if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of $Y_1$. Applying Bayes rule:

$$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$

So you have the exponential distribution $p(Y_1)$ that we already understand, and the other stranger thing $p(emails_1 = 1|Y_1)$ which is the probability of not having any other mail sent after the first one, until time 1, given $Y_1$. It is equal to $\int_1^\infty p(Y_2|Y_1) dY_2$ and it raises as $Y_1$ gets closer to 1. Actually, as the integral of an exponential function is exponential as well, it turns out, with few passages, that $p(emails_1 = 1|Y_1) = exp(1-Y_1)$, so:

$$p(Y_1|emails_1 = 1) \propto exp(Y_1)exp(1-Y_1) \propto 1$$

where $exp$ is the esponential distribution density, whatever is the (equal) rate parameter. Last passage is easily verifiable from multiplying those exponential densities.

Your updated distribution of $Y_1$ is uniform, and expected value can be immediately derived.

If you'd like a much shorter, more intuitive explanation, mind that Poisson process is symmetrical, and that the distribution of $Y_1$ is the same looked from 0 and from 1, so of course its expected value must be 0.5. This explanation doesn't account for the rest of its distribution though.

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carlo
carlo
  • 5.1k
  • 1
  • 14
  • 31

This is actually a bayesian problem. Time $Y_1$ if you don't know the number of mails at time 1, is exponentially distributed, you got that right. But when you get the additional information that at time 1 Alice only sent one e-mail, you have to update your distribution of $Y_1$. Applying Bayes rule:

$$p(Y_1|emails_1 = 1) \propto p(emails_1 = 1|Y_1)p(Y_1).$$

So you have the exponential distribution $p(Y_1)$ that we already understand, and the other stranger thing $p(emails_1 = 1|Y_1)$ which is the probability of not having any other mail sent after the first one, until time 1, given $Y_1$. It is equal to $\int_1^\infty p(Y_2|Y_1) dY_2$ and it raises as $Y_1$ gets closer to 1. Actually, as the integral of an exponential function is exponential as well, it turns out, with few passages, that $p(emails_1 = 1|Y_1) = p(1-Y_1)$ so:

$$p(Y_1|emails_1 = 1) \propto exp(Y_1)exp(1-Y_1) \propto 1$$

where $exp$ is the esponential distribution density, whatever is the (equal) rate parameter. Last passage is easily verifiable from multiplying those exponential densities.

Your updated distribution of $Y_1$ is uniform, and expected value can be immediately derived.

If you'd like a much shorter, more intuitive explanation, mind that Poisson process is symmetrical, and that the distribution of $Y_1$ is the same looked from 0 and from 1, so of course its expected value must be 0.5. This explanation doesn't account for the rest of its distribution though.