- #1
Mogarrr
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Today in class, there was an example where I didn't understand certain justifications. The example goes something like this:
A casino runs a game of chance where you toss a coin and they pay $1 if you get heads , and you pay $1 if you get tails. The coin is a fair coin.
A gambler starts with fortune z. The goals is to increase the fortune to g≥z. Said gambler plays until they reach g or their fortune is $0.
Let A be the event of reaching the goal g starting from fortune z.
Let Pr(A) = ψ(z).
It's clear that ψ(0)=0 and ψ(g)=1, by the rules the gambler uses. We want a formula ψ(z) for 0<z<g. We use the law of total probability, conditioning on the first toss.
Let [itex]B_1[/itex]=first toss is heads
Let [itex]B_2[/itex]=first toss is tails, as this forms a partition of the sample space.
Pr(A)=[itex]Pr(B_1)\cdot Pr(A|B_1) + Pr(B_2)\cdot Pr(A|B_2) [/itex], so
ψ(z) = (1/2)ψ(z+1) + (1/2)ψ(z-1) = (1/2)(ψ(z+1)+ψ(z-1))
and this is where I have a big question mark over my head
ψ(z) is a linear function of z
[itex] \exists a,b: ψ(z)=az+b [/itex]
[itex] \psi (0)=0 \Rightarrow b=0 [/itex]
[itex] \psi (g) =1 \Rightarrow a = \frac 1{g} [/itex]
[itex] \Rightarrow \psi (z)= \frac {z}{g}, \forall z: 0 \leq z \leq g [/itex]
...So how is it justified that the recursive function is a linear function. The professor talked about a linear interpolation happening, but I just don't see how you can make that jump. I can follow all other lines of reasoning, so I just want to talk about this.
Anybody here know how that step is justified?
A casino runs a game of chance where you toss a coin and they pay $1 if you get heads , and you pay $1 if you get tails. The coin is a fair coin.
A gambler starts with fortune z. The goals is to increase the fortune to g≥z. Said gambler plays until they reach g or their fortune is $0.
Let A be the event of reaching the goal g starting from fortune z.
Let Pr(A) = ψ(z).
It's clear that ψ(0)=0 and ψ(g)=1, by the rules the gambler uses. We want a formula ψ(z) for 0<z<g. We use the law of total probability, conditioning on the first toss.
Let [itex]B_1[/itex]=first toss is heads
Let [itex]B_2[/itex]=first toss is tails, as this forms a partition of the sample space.
Pr(A)=[itex]Pr(B_1)\cdot Pr(A|B_1) + Pr(B_2)\cdot Pr(A|B_2) [/itex], so
ψ(z) = (1/2)ψ(z+1) + (1/2)ψ(z-1) = (1/2)(ψ(z+1)+ψ(z-1))
and this is where I have a big question mark over my head
ψ(z) is a linear function of z
[itex] \exists a,b: ψ(z)=az+b [/itex]
[itex] \psi (0)=0 \Rightarrow b=0 [/itex]
[itex] \psi (g) =1 \Rightarrow a = \frac 1{g} [/itex]
[itex] \Rightarrow \psi (z)= \frac {z}{g}, \forall z: 0 \leq z \leq g [/itex]
...So how is it justified that the recursive function is a linear function. The professor talked about a linear interpolation happening, but I just don't see how you can make that jump. I can follow all other lines of reasoning, so I just want to talk about this.
Anybody here know how that step is justified?