<p>I haven't read the paper, and I think the expression is only an approximation, but the basic idea is that this is an <a href="https://en.wikipedia.org/wiki/Poisson_point_process#Inhomogeneous_Poisson_point_process" rel="nofollow noreferrer">inhomogeneous Poisson process.</a></p>

<p>In order for a fork to occur, some unaware node needs to find a block. This is 1 minus the probability that no unaware node will ever find a block. Which itself means that no unaware node finds a block in 1st second, no unaware node find a block in 2nd second, etc. This probability is the product of the probabilities of the individual seconds (since they are all independent).</p>

<p>Each individual probability is roughly (1-P_b)^(1-f(t)) (if f(t)=0 then all nodes are unaware and the probability is the same as no block being found at all a block at all, if f(t)=1 then all nodes are aware and certainly no block will be found by an unaware node, and in between it interpolates. If it's not clear why the interpolation is via a power, keep in mind that for small P_b, the power is irrelevant since the expression is roughly 1-P_b*(1-f(t)).</p>

<p>The combined probability, as mentioned, is the product of probabilities, which is like (1-P_b) to the power of a sum of (1-f(t)), which is like (1-P_b) to the power of an integral.</p>

<p>If it was up to me I would have written the expression differently, but these are the general ideas.</p>

CC BY-SA 4.0
2018-05-02T23:42:18.450Z
That makes a lot of sense now given that the process of finding blocks is independent. In your answer, do you mean to write the exponential term _with_ the integral, as in the original expression? (I know there's no support for MathJax on here unlike MSE; just making sure).
an4s
CC BY-SA 4.0
2018-05-03T09:28:14.070Z
@an4s: Actually, there's a paragraph I simply forgot to write which addresses the integral. Added it now. An integral is like a sum, and (1-P_b) to the power of a sum is like a product of terms, each of which is (1-P_b) raised to some power.

This was a bit handwavy but may make more sense by reading up on exponential distribution, Poisson process so on. And I'll mention again that to me the expression seems only approximate, (1-P_b) should be replaced with exp(-P_b) where P_b is the block finding rate.
Meni Rosenfeld
CC BY-SA 4.0
2018-05-02T22:48:17.160Z
2018-05-03T09:24:53.120Z
2018-05-03T09:24:53.120Z
Meni Rosenfeld
<p>I am reading <a href="https://ieeexplore.ieee.org/abstract/document/6688704" rel="nofollow noreferrer">this</a> paper on how information is propagated on the Bitcoin network. The authors present a model to predict the rate at which forks occur in the blockchain which is given as below
<a href="https://i.sstatic.net/CcZZX.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CcZZX.png" alt="enter image description here"></a>
Here, <em>F</em> is the number of conflicting blocks in the network, <em>Pb</em> is the probability of the network finding a block <em>b</em> in a given second (ideally, 1/600 since a block is expected every 10 mins = 600 secs) and <em>f(t)</em> represents the ratio of nodes that hear about the block <em>b</em> in <em>t</em> secs. However, what I cannot understand is how this entire expression is derived. I understand that the term in the exponent represents the mean amount of time it takes for the network to learn about a block (and this value can be derived from a graph given in the paper). I assume <em>1 - Pb</em> represents the probability of the network finding more blocks in the remaining 599 seconds in the 10-minute interval. Why is this probability raised to the mean amount of time it takes for the network to learn about a block?</p>

<p>Any explanation would be appreciated.</p>

<blockchain><blockchain-fork>
Explanation of probability of forks in the Bitcoin blockchain
an4s
# Explanation of probability of forks in the Bitcoin blockchain

(Asked by: username_0 on 2018-05-02)

I am reading [this](https://ieeexplore.ieee.org/abstract/document/6688704) paper on how information is propagated on the Bitcoin network. The authors present a model to predict the rate at which forks occur in the blockchain which is given as below
[![enter image description here](https://i.sstatic.net/CcZZX.png)](https://i.sstatic.net/CcZZX.png)
Here, *F* is the number of conflicting blocks in the network, *Pb* is the probability of the network finding a block *b* in a given second (ideally, 1/600 since a block is expected every 10 mins = 600 secs) and *f(t)* represents the ratio of nodes that hear about the block *b* in *t* secs. However, what I cannot understand is how this entire expression is derived. I understand that the term in the exponent represents the mean amount of time it takes for the network to learn about a block (and this value can be derived from a graph given in the paper). I assume *1 - Pb* represents the probability of the network finding more blocks in the remaining 599 seconds in the 10-minute interval. Why is this probability raised to the mean amount of time it takes for the network to learn about a block?

Any explanation would be appreciated.

**Tags:** blockchain, blockchain-fork


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## 1 answers

### Answer by username_1 on 2018-05-02. Score: 2 | Accepted answer

I haven't read the paper, and I think the expression is only an approximation, but the basic idea is that this is an [inhomogeneous Poisson process.](https://en.wikipedia.org/wiki/Poisson_point_process#Inhomogeneous_Poisson_point_process)

In order for a fork to occur, some unaware node needs to find a block. This is 1 minus the probability that no unaware node will ever find a block. Which itself means that no unaware node finds a block in 1st second, no unaware node find a block in 2nd second, etc. This probability is the product of the probabilities of the individual seconds (since they are all independent).

Each individual probability is roughly (1-P\_b)^(1-f(t)) (if f(t)=0 then all nodes are unaware and the probability is the same as no block being found at all a block at all, if f(t)=1 then all nodes are aware and certainly no block will be found by an unaware node, and in between it interpolates. If it's not clear why the interpolation is via a power, keep in mind that for small P\_b, the power is irrelevant since the expression is roughly 1-P\_b\*(1-f(t)).

The combined probability, as mentioned, is the product of probabilities, which is like (1-P\_b) to the power of a sum of (1-f(t)), which is like (1-P\_b) to the power of an integral.

If it was up to me I would have written the expression differently, but these are the general ideas.

> **Comments:**
> **username_0**: That makes a lot of sense now given that the process of finding blocks is independent. In your answer, do you mean to write the exponential term _with_ the integral, as in the original expression? (I know there's no support for MathJax on here unlike MSE; just making sure).  
> **username_1**: @username_0: Actually, there's a paragraph I simply forgot to write which addresses the integral. Added it now. An integral is like a sum, and (1-P_b) to the power of a sum is like a product of terms, each of which is (1-P_b) raised to some power.  This was a bit handwavy but may make more sense by reading up on exponential distribution, Poisson process so on. And I'll mention again that to me the expression seems only approximate, (1-P_b) should be replaced with exp(-P_b) where P_b is the block finding rate.  

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