Sir Ronald Fisher, the stochastic frog (Frogus Stokastikus) is in trouble. Stuck in an infinite well, his energy is depleted. For each minute of effort, he finds that he either climbs up one brick or falls down one brick. What can we say about his (probabilistic) plight?
The below video was originally produced as a Seren Network Masterclass organised by the Port Neath Talbot, Powys and Bridgend Hub.
Test your understanding of the video with the following.
1. What is the probability that the frog is at a height of 4 after 8 time steps? (i.e. what is P(H(8)=4)?)
For this we need 6 up turns and 2 down turns so: \(P(H(8)=4) = {^8 C _4} (\frac{1}{2})^6(\frac{1}{2})^2\)
\(\frac{28}{256} = \frac{7}{64}\)
2. What is the probability that the frog ends up lower than he started after 6 time steps?
We want \(P(H(6)<0) = P(H(6)=-2) + P(H(6)=-4) +P(H(6)=-6)\)
For -2 we need 2 up turns and 4 down. For -4 we need 1 up turn and 5 down turns. For -6 we need 6 down turns.
so: \(P(H(6)<0) = {^6 C _2} (\frac{1}{2})^2(\frac{1}{2})^4 + {^6 C _1} (\frac{1}{2})^1(\frac{1}{2})^5 + {^6 C _0} (\frac{1}{2})^6\)
\(\frac{15 + 6 + 1}{64} = \frac{22}{64} = \frac{11}{32}\)
3. Given that the frog is at a height of 2 after 4 time steps, what is the probability that he is at a height of 3 after 7 steps?
We are effectively moving the 0 point up to 2 and setting t=4 to t=0. This is equivalent to \(P(H(7-4)=3-2) = P(H(3)=1)\)
\(\frac{3}{8}\)