Help me understand this: $ Pr(T < t \mid y) = \int_{0}^{\infty }\int_{\Omega }k(x)p(t, x\mid y)\,dx\,dt \ $? How do I read this equation (especially the left side) in terms of a Continuous Markov Process model? 
$$
Pr(T < t \mid y) = \int_{0}^{\infty }\int_{\Omega }k(x)p(t, x\mid y)\,dx\,dt \\
$$
Where $ p(t,x) $ is the transition probability density of a diffusion process, and $ y $, I assume, is the previous state of $ x $.
Is the left hand side strictly a function of $ t $? I suspect yes but...
Is the $ p( \  \mid y) $ notation necessary ?
Any help would be greatly appreciated. Thank you. 
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Here is an example: 
Diffusion Equation: 
$$ dx=\mu dt + \sigma dW \ Eq. 1\\ $$
PDF of the diffusion:
$$ p(t,x)=\frac{e^{-\frac{(x-t\mu )^{2}}{2\sigma t^{2}}}}{\sqrt{2\pi }\sqrt{t\sigma ^{2}}} \ Eq. 2 \\ $$
Killing function: 
$$ k(x)=\lambda x^{2} \ Eq. 3$$
$$ k(x)=\lambda x^{2}\Rightarrow T = \lambda x^{2}\Rightarrow x = \sqrt{\frac{t}{\lambda }} \\ $$
The original equations PDF (Integrated w/Mthematica):
$$ Pr(T < t, \tau  > T \mid y) =
\int_{0}^{t}\int_{\Omega }k(x)p(t, x)\,dx\,dt =
\frac{1}{6}\lambda \sqrt{\frac{1}{\sigma ^{2}t}}t^{2}\sqrt{\sigma ^{2}t}(3\sigma ^{2}+2\mu ^{2}t) \ Eq. 4 \\ $$
For some arbitrary values:
$$ \mu=0.05, \ \sigma=0.5 $$
Plotting with $ \lambda=0.5 $: 


Plotting with $ \lambda=0.01 $:


{Axes: t=horizontal, x=vertical}
So, when $ \lambda $ is larger - more paths are being killed; hence, "LHS CDF" increases much faster. 
Nevertheless, the LHS PDF fully integrated does not equal 1 so its not a distribution. How to make sense of all this?
 A: This is a conditional probability. You should try to understand the left part before the right part.
In the left part, it says "what's the probability of your random variable t is greater than a constant T given that you know the previous states
For example, if you want to predict the probability next raining day after next weekend and you know the history of weather. y would be the information of the weather data, t would be the random variable for the next raining day and T is the next weekend. The (|y) information is absolutely essential, your predicted probability of the next raining day given that you know the history is not the same as your prediction without knowing the history. For example, you can safely assume the probability is zero if you live in the drought area.
Unless you're comfortable with the concept of conditional probability, I wouldn't be too concern with the right part of the formula. Essentially it says "integrate and sum up all the probability over the two dimensions: time and states".
A: I would read the left side of that equation as "the probability that the random variable T is less than the specific value t (which is in the domain of T) given that you know y." From what I remember, the left side is actually a function of y because, as the right side of the equation indicates, the t's are integrated out.
Maybe, in terms of the Markov Process, read the T as the time or steps to convergence given that you are at time y, or that you have come y steps. Notice that the domain of t on the right side is from zero to infinity. So, by integrating over all future times or steps, you are finding the probability that you converge before t over all the states available in the domain of x.
