Shipping If we consider the amount of water added to the larger bucket v v v v = t v=t v = t
Using this we can find E [ X v ] = m ( v ) E[X_{v}] = m(v) E [ X v ] = m ( v ) X v X_{v} X v v = 1 v=1 v = 1 E [ X v + 1 ] = E [ X v ] + 1 \mathbb{E}[X_{v}+1]=\mathbb{E}[X_{v}]+1 E [ X v + 1 ] = E [ X v ] + 1
Now we can find E [ X v ] = m ( v ) = F ( v ) + ∫ 0 v m ( v − s ) f ( s ) d s E[X_{v}] = m(v) = F(v)+\int_{0}^{v}m(v-s)f(s)ds E [ X v ] = m ( v ) = F ( v ) + ∫ 0 v m ( v − s ) f ( s ) d s v ∈ [ 0 , 1 ] v\in[0,1] v ∈ [ 0 , 1 ]
Thus we get m ( v ) = v + ∫ 0 v m ( v − s ) d s m(v)=v+\int_{0}^{v}m(v-s)ds m ( v ) = v + ∫ 0 v m ( v − s ) d s F ( v ) = v , v ∈ [ 0 , 1 ] F(v)=v, v\in[0,1] F ( v ) = v , v ∈ [ 0 , 1 ] f ( s ) = 1 f(s)=1 f ( s ) = 1
Now let u = v − s ⟹ d u = − d s u=v-s \implies du=-ds u = v − s ⟹ d u = − d s
Thus we get
m ( v ) = v + ∫ 0 v m ( v − s ) d s = v − ∫ v 0 m ( u ) d u ⟹ m ( v ) = v + ∫ 0 v m ( u ) d u ⟹ m ′ ( v ) = 1 + m ( v ) + m ( 0 ) = 1 + m ( v ) Since m ( 0 ) = 0 ⟹ m ′ ( v ) − m ( v ) = 1 ⟹ d d v ( e ∫ − 1 d v m ( v ) ) = e − v Using integrated factor e − v m ( v ) = ∫ e − v d v = − e − v + C m ( v ) = − 1 + C e − v Since m ( 0 ) = 0 , C = 1 ∴ m ( v ) = e v − 1 For v ∈ [ 0 , 1 ] \begin{align*}
m(v) & = v+\int_{0}^{v}m(v-s)ds \\
& =v-\int_{v}^{0}m(u)du \\
\implies m(v) & = v+\int_{0}^{v}m(u)du \\
\implies m'(v) & =1+m(v)+m(0) \\
& =1+m(v) & \text{Since }m(0)=0 \\
\implies m'(v)-m(v) & =1 \\
\implies \frac{d}{dv}\left( e^{\int_{}^{}-1dv}m(v) \right) & =e^{-v} & \text{Using integrated factor} \\
e^{-v}m(v) & =\int_{}^{}e^{-v}dv \\
& =-e^{-v}+C \\
m(v) & =-1+\frac{C}{e^{-v}} & \text{Since }m(0)=0, C=1 \\
\therefore m(v) & =e^{v}-1 & \text{For }v\in[0,1]
\end{align*} m ( v ) ⟹ m ( v ) ⟹ m ′ ( v ) ⟹ m ′ ( v ) − m ( v ) ⟹ d v d ( e ∫ − 1 d v m ( v ) ) e − v m ( v ) m ( v ) ∴ m ( v ) = v + ∫ 0 v m ( v − s ) d s = v − ∫ v 0 m ( u ) d u = v + ∫ 0 v m ( u ) d u = 1 + m ( v ) + m ( 0 ) = 1 + m ( v ) = 1 = e − v = ∫ e − v d v = − e − v + C = − 1 + e − v C = e v − 1 Since m ( 0 ) = 0 Using integrated factor Since m ( 0 ) = 0 , C = 1 For v ∈ [ 0 , 1 ] Which means E [ X 1 + 1 ] = e 1 − 1 + 1 = e \mathbb{E}[X_{1}+1] = e^{1}-1+1=e E [ X 1 + 1 ] = e 1 − 1 + 1 = e
To find E [ X 2 + 1 ] = m ( 2 ) + 1 \mathbb{E}[X_{2}+1]=m(2)+1 E [ X 2 + 1 ] = m ( 2 ) + 1 v = 1 v=1 v = 1
m ( v ) = F ( v ) + ∫ 0 v m ( v − s ) f ( s ) d s for v ∈ [ 1 , 2 ] since f ( v ) = 0 ≥ 1 and F ( v ) = 1 , ∀ v ≥ 1 m ( v ) = 1 + ∫ 0 1 m ( v − s ) d s = 1 + ∫ v v − 1 − m ( u ) d u = 1 + ∫ v − 1 v m ( u ) d u m ′ ( u ) = m ( v ) − m ( v − 1 ) since v − 1 ∈ [ 0 , 1 ] then m ( v − 1 ) = e v − 1 − 1 = m ( v ) − ( e v − 1 ) m ′ ( v ) − m ( v ) = 1 − e v − 1 Using integrated factors d d v ( e − v m ( v ) ) = e − v − e − 1 e − v m ( v ) = − e − v − v e − 1 + C m ( v ) = − 1 − v e v − 1 + C e v Since m ( 1 ) = e − 1 m ( v ) = e + 1 e e v − 1 − v e v − 1 , v ∈ [ 1 , 2 ] \begin{align*}
m(v) & =F(v)+\int_{0}^{v}m(v-s)f(s)ds & \text{for } v\in[1,2] \\
\text{since } & f(v) =0\geq 1 \text{ and }F(v)=1, \forall v\geq 1 \\
m(v) & =1+\int_{0}^{1}m(v-s)ds \\
& =1+\int_{v}^{v-1}-m(u)du \\
& =1+\int_{v-1}^{v}m(u)du \\
m'(u) & =m(v)-m(v-1) \\
\text{since } & v-1\in[0,1] \text{ then }m(v-1)=e^{v-1}-1 \\
& =m(v)-(e^{v}-1) \\
m'(v)-m(v) & =1-e^{v-1} \\
& \text{ Using integrated factors} \\
\frac{d}{dv}(e^{-v}m(v)) & = e^{-v}-e^{-1} \\
e^{-v}m(v) & =-e^{-v}-ve^{-1}+C \\
m(v) & =-1-ve^{v-1}+Ce^{v} \\
\text{Since } & m(1)=e-1 \\
m(v) & =\frac{e+1}{e}e^{v}-1-ve^{v-1}, v\in[1,2]
\end{align*} m ( v ) since m ( v ) m ′ ( u ) since m ′ ( v ) − m ( v ) d v d ( e − v m ( v )) e − v m ( v ) m ( v ) Since m ( v ) = F ( v ) + ∫ 0 v m ( v − s ) f ( s ) d s f ( v ) = 0 ≥ 1 and F ( v ) = 1 , ∀ v ≥ 1 = 1 + ∫ 0 1 m ( v − s ) d s = 1 + ∫ v v − 1 − m ( u ) d u = 1 + ∫ v − 1 v m ( u ) d u = m ( v ) − m ( v − 1 ) v − 1 ∈ [ 0 , 1 ] then m ( v − 1 ) = e v − 1 − 1 = m ( v ) − ( e v − 1 ) = 1 − e v − 1 Using integrated factors = e − v − e − 1 = − e − v − v e − 1 + C = − 1 − v e v − 1 + C e v m ( 1 ) = e − 1 = e e + 1 e v − 1 − v e v − 1 , v ∈ [ 1 , 2 ] for v ∈ [ 1 , 2 ] Thus, m ( 2 ) = e + 1 e e 2 − 1 − v e 1 = e 2 − e − 1 m(2)=\frac{e+1}{e}e^{2}-1-ve^{1}=e^{2}-e-1 m ( 2 ) = e e + 1 e 2 − 1 − v e 1 = e 2 − e − 1 E [ X 2 + 1 ] = m ( 2 ) + 1 = e 2 − e \mathbb{E}[X_{2}+1]=m(2)+1=e^{2}-e E [ X 2 + 1 ] = m ( 2 ) + 1 = e 2 − e
All the states form a single recurrence class since every state communicates with state m + 1 = 6 m+1=6 m + 1 = 6 C 1 = { 1 , 2 , 3 , 4 , 5 , 6 } C_{1}=\left\{ 1,2,3,4,5,6 \right\} C 1 = { 1 , 2 , 3 , 4 , 5 , 6 } C 1 C_{1} C 1
r 6 , 6 ( 1 ) = 0 r_{6,6}(1)=0 r 6 , 6 ( 1 ) = 0 r 6 , 6 ( 2 ) > 0 r_{6,6}(2)>0 r 6 , 6 ( 2 ) > 0 6 6 6 r 6 , 6 ( 3 ) > 0 r_{6,6}(3)>0 r 6 , 6 ( 3 ) > 0 n ∈ [ 1 , 5 ] n\in[1,5] n ∈ [ 1 , 5 ] 6 → n → n + 1 ( mod 5 ) → 6 6\to n\to n+1(\text{mod } 5) \to 6 6 → n → n + 1 ( mod 5 ) → 6
However, since gcd ( 2 , 3 ) = 1 \texttt{gcd}(2,3)=1 gcd ( 2 , 3 ) = 1
Let's consider the the solution to the equations
π = π [ 0 1 3 0 0 1 3 1 3 1 3 0 1 3 0 0 1 3 0 1 3 0 1 3 0 1 3 0 0 1 3 0 1 3 1 3 1 3 0 0 1 3 0 1 3 1 5 1 5 1 5 1 5 1 5 0 ] , ∑ j = 1 m π j = 1 \mathbf{\pi}=\mathbf{\pi}\begin{bmatrix}
0 & \frac{1}{3} & 0 & 0 & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & 0 & \frac{1}{3} & 0 & 0 & \frac{1}{3} \\
0 & \frac{1}{3} & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\
0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} \\
\frac{1}{3} & 0 & 0 & \frac{1}{3} & 0 & \frac{1}{3} \\
\frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0
\end{bmatrix}, \sum_{j=1}^{m}\pi _{j}=1 π = π 0 3 1 0 0 3 1 5 1 3 1 0 3 1 0 0 5 1 0 3 1 0 3 1 0 5 1 0 0 3 1 0 3 1 5 1 3 1 0 0 3 1 0 5 1 3 1 3 1 3 1 3 1 3 1 0 , j = 1 ∑ m π j = 1 where π = ( π 1 , … π 6 ) \mathbf{\pi}=\left( \pi_{1},\dots \pi_{6} \right) π = ( π 1 , … π 6 )
Using symmetry we can see that π 1 = π 2 = π 3 = π 4 = π 5 = x \pi_{1}=\pi_{2}=\pi_{3}=\pi_{4}=\pi_{5}=x π 1 = π 2 = π 3 = π 4 = π 5 = x 5 x + π 6 = 1 5x+\pi_{6}=1 5 x + π 6 = 1 π = ( 3 20 , 3 20 , 3 20 , 3 20 , 3 20 , 1 4 ) \mathbf{\pi}=\left(\frac{3}{20}, \frac{3}{20}, \frac{3}{20}, \frac{3}{20}, \frac{3}{20}, \frac{1}{4} \right) π = ( 20 3 , 20 3 , 20 3 , 20 3 , 20 3 , 4 1 )
The transition rates of of q i j q_{ij} q ij
For i ∈ [ 1 , m ] i\in[1,m] i ∈ [ 1 , m ] The transition rates from any of these states is λ \lambda λ λ \lambda λ
q i j = { λ if j = m + 1 0 otherwise q_{ij}=\begin{cases}
\lambda & \text{if }j=m+1 \\
0 & \text{otherwise}\end{cases} q ij = { λ 0 if j = m + 1 otherwise For i = m + 1 i=m+1 i = m + 1 q i j = { 0 if j = m + 1 λ m otherwise q_{ij}=\begin{cases}
0 & \text{if }j=m+1 \\
\frac{\lambda}{m} & \text{otherwise}
\end{cases} q ij = { 0 m λ if j = m + 1 otherwise Thus we have
Q = ( 0 0 … 0 λ 0 0 … 0 λ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 … 0 λ λ m λ m … λ m 0 ) Q=\begin{pmatrix}
0 & 0 & \dots & 0 & \lambda \\
0 & 0 & \dots & 0 & \lambda \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 0 & \lambda \\
\frac{\lambda}{m} & \frac{\lambda}{m} & \dots & \frac{\lambda}{m} & 0
\end{pmatrix} Q = 0 0 ⋮ 0 m λ 0 0 ⋮ 0 m λ … … ⋱ … … 0 0 ⋮ 0 m λ λ λ ⋮ λ 0 And the rates out of each state is λ \lambda λ
P = ( 0 0 … 0 1 0 0 … 0 1 ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 … 0 1 1 m 1 m … 1 m 0 ) P=\begin{pmatrix}
0 & 0 & \dots & 0 & 1 \\
0 & 0 & \dots & 0 & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 0 & 1 \\
\frac{1}{m} & \frac{1}{m} & \dots & \frac{1}{m} & 0
\end{pmatrix} P = 0 0 ⋮ 0 m 1 0 0 ⋮ 0 m 1 … … ⋱ … … 0 0 ⋮ 0 m 1 1 1 ⋮ 1 0 Let
A = ( − λ 0 … 0 λ 0 − λ … 0 λ ⋮ ⋮ ⋱ ⋮ ⋮ 0 0 … − λ λ λ m λ m … λ m − λ ) A =\begin{pmatrix}
-\lambda & 0 & \dots & 0 & \lambda \\
0 & -\lambda & \dots & 0 & \lambda \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & -\lambda & \lambda \\
\frac{\lambda}{m} & \frac{\lambda}{m} & \dots & \frac{\lambda}{m} & -\lambda
\end{pmatrix} A = − λ 0 ⋮ 0 m λ 0 − λ ⋮ 0 m λ … … ⋱ … … 0 0 ⋮ − λ m λ λ λ ⋮ λ − λ Then the backwards equations are
P ′ ( t ) = A P ( t ) P'(t)=AP(t) P ′ ( t ) = A P ( t ) Therefore, to solve
Therefore, we are done □ \square □
