This question is about the poison telegraphic process, which is a process with a random variable and AT T and the way the process works is that at time zero and that he is assigned either plus one or minus one randomly, with probability, 50% each. And then events occur according to a post on process, and each time an event occurs, the parody of an itty switches. So, for example, if entity is plus one and an event occurs, then it switches to minus one and vice versa. We're also told that the probability of an even number of events occurring in the time interval from zero to t is 0.5. And so for part, A were asked to explain why for time greater than zero, given this statement here were asked to explain why the following is true. So if we look at the first of these two, it's explained why. The conditional probability that n A T equals plus one given in at zero is equal to plus one equals p. And remember, P is the probability of an even number of events in time zero to t. So if we think of our time interval. Let's say T is here. And as events occur according to a plus on process, that same event happens here. That's we'll call this capital T someone for the first event t's up to for the second and so on. So we could say that T I is the I've event in the interval from zero to t. So then if and at zero is plus one then we have and at the time of the first event will be minus one and at the time of the second event, plus one, which means that N a. T is positive Onley for even I so for an even number of events. So that and that t will be plus one on Lee if we haven't even number of events. So this means that the probability of enmity being plus one given that we started and at zero is plus one is simply this probability here the probability of the statement being true which we know to be p the probability of an even number of events in zero to t. And now for the second statement, we can use the same logic. So this time if and that zero is equal to minus one. That's the condition then and at the time of the first event is it would have plus one. And at the time of the second event is minus one and so on. And therefore you can see that n a. T is plus one Onley when the number of events on the interval is odd numbered. You can say that if we start with an at zero equal to minus one in a T is equal to plus one only if there is an odd number of events on the time interval from zero to t. So therefore, we could say that the probability that entity is equal to minus one alright rather plus one given that we started and at zero equal to minus one is the truth of this statement and the probability of an odd number of events is one minus p. Remember that P is the probability of an even number of events. So therefore we have we have explained why these statements in part a are true. Uh huh. So now, for part B were asked to use our results from part A and the law of total probability to show that the probability that entity is equal to plus one is equal to 0.5. We're all time greater than or equal to zero. Using the law of total probability probability that entity equals plus one is expressed as the probability that entity it was, plus one given at zero. It was one times the probability that and it's zero equals one. And so this is the law of total probability because we're finding the probability of n a t one plus one conditioned on an exhaustive partition of and at zero so and at zero can only take on two values, either minus one or plus one. And so, if we condition it on both of those values or all partitions of an ID zero in this manner and multiplied by the probability of in at zero equaling that value, we arrive at the total probability for entity equaling plus one. One more thing that we should note here is that probability of an it zero equaling plus one and the probability of an 80 equaling minus one are both have, and that's by the definition of the Poisson telegraphic process, and therefore that gives us the probability of entity being plus one equal to so this probability and this. So that's this probability and this probability are both half. And this probability and this probability are the results from part A of this question. So it's straightforward to find the answer. At this point, we have half times P plus half times one minus p, which comes out to have then for part C. We're asked to establish the property. That function mean of a telegraphic process is equal to zero, and standard deviation of the process is equal to one. So we have. This is equal to the expected value, the telegraphic process, random variable. And this is equal to entity times the probability of entity for all entity. And so there are only two entities. It's either plus one or minus one. So there's plus one and we've established that the probability of Equalling plus one his half and it can also equal minus one with probability one half it was zero. Now we can write the variance of energy, So an a t you could be minus one minus the main, which is zero times the probability that entity equals minus one, and it could be plus one. So one minus the mean squared times the probability that entity takes on plus one. And this comes out to you one times one half plus one times one half equals one. And therefore, we can say that the standard deviation of the process is also equal to one because it is the square root of one.