Suppose that a model has four states: $0, 1, 2,$ and $3$, and the only possible transitions between these states is $0\rightarrow 1$, $0\rightarrow 2$, and $0\rightarrow 3$. For $t\geq 0$, $\mu_{x+t}^{01} = 0.3$, $\mu_{x+t}^{02} = 0.5$ and $\mu_{x+t}^{03} = 0.7$. Compute $p_x^{02}$.
Now, suppose instead we have a model with three states: $0, 1$ and $2$, with the only possible transitions being $0\rightarrow 1$ and $0\rightarrow 2$ (ie a double decrement model). Let $\mu_x^{01} = a$ and $\mu_x^{02} = b$ for $0\leq x\leq 1$. Determine expressions for $_1p_x^{00}$, $_1p_x^{01}$ and $_1p_x^{02}$. If $b = ka$, show that
$$_1p_x^{01} = \frac{1 - _1p_x^{00}}{k+1}$$
My attempt:
For the first part:
$$_tp_x^{00} = _tp_x^{\bar{00}} = e^{-\int_{0}^{t} \mu_{x+r}^{01} + \mu_{x+r}^{02} + \mu_{x+r}^{03} \,dr} = e^{-\int_{0}^{t} 1.5 \,dr} = e^{-\frac{3t}{2}}$$
$$_tp_x^{11} = _tp_x^{\bar{11}} = e^0 = 1$$
So
$$p_x^{02} = \int_{0}^{1} tp_x^{00} \times \mu_{x+t}^{02} \times _{1-t}p_{x+t}^{22} \,dt = 0.26$$
,which follows from the chain rule.
For the second part:
$$_1p_x^{00} = _1p_x^{\bar{00}} = e^{-\int_{0}^{1} \mu_{x+r}^{01} + \mu_{x+r}^{02} \,dr} = e^{-\int_{0}^{1} a + b \,dr} = e^{-(a+b)}$$
Then
$$_tp_x^{00} = _tp_x^{\bar{00}} = e^{-\int_{0}^{t} \mu_{x+r}^{01} + \mu_{x+r}^{02} \,dr} = e^{-\int_{0}^{t} a + b \,dr} = e^{-(a+b)t}$$
$$_tp_x^{11} = _tp_x^{\bar{11}} = e^0 = 1$$
$$_1p_x^{01} = \int_{0}^{1} tp_x^{00} \times \mu_{x+t}^{01} \times _{1-t}p_{x+t}^{11} \,dt = \frac{a(1-e^{-(a+b)})}{a+b}$$
Similarly
$$_tp_x^{22} = _tp_x^{\bar{22}} = e^0 = 1$$
$$_1p_x^{02} = \int_{0}^{1} tp_x^{00} \times \mu_{x+t}^{02} \times _{1-t}p_{x+t}^{22} \,dt = \frac{b(1-e^{-(a+b)})}{a+b}$$
Then with $b=ka$, we get that
$$_1p_x^{01} = \frac{a(1-e^{-(a+b)})}{a+b} = \frac{a(1-e^{-(a+b)})}{a+ka} = \frac{(1-e^{-(a+b)})}{1+k} = \frac{1 - _1p_x^{00}}{1+k}$$
Is what I have done correct? Any assistance is much appreciated.