Probability virus question at different infection rates

In summary: The first thing that has to be resolved is the interpretation of the question. Do you read it that the recovery time starts immediately on being infected (so recovery may occur without ever becoming infectious) or on becoming infectious?The first thing that has to be resolved is the interpretation of the question. Do you read it that the recovery time starts immediately on being infected (so recovery may occur without ever becoming infectious) or on becoming infectious?
  • #1
Mark53
93
0

Homework Statement


From various studies, it is known that once an individual is infected with a virus, they become infectious at rate λ. The individual will recover at rate λ, independent of the time it took for them to become infectious. Let X be the total amount of time an individual has this virus.

(a) What is the distribution of X?
(b) What is the probability an infected individual has this virus for less than 1 day?
(c) If λ = 1, what is the probability that after getting the virus, they become infectious in less than one day and then they recover in less than one day?

The Attempt at a Solution


a)
X~Poi(λ)

b)

P(x<1)=P(x=0)

=e^(-λ)λ^0/0!
=e^(-λ)

c)

P(x<1)nP(x<1)
=P(x=0)*P(x=0)
=e^(-λ)*e^(-λ)
=e^(-2)

not sure if this part is correct
 
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  • #2
I wonder if you have misread the question. X covers the time both to become infectious and then to recover. Each of those has an expected duration of 1/λ. You answer to a) would give E(X)=1/λ.
 
  • #3
Mark53 said:

Homework Statement


... at rate λ... Let X be the total amount of time an individual has this virus.

(a) What is the distribution of X?

The Attempt at a Solution


a)
X~Poi(λ)

I'm not totally happy with the way this problem is worded. Keep in mind that a Poisson is a discrete distribution (or counting process) characterized by exponentially distributed inter-arrival times. ##X##, here is the total amount of time a given person has the virus (clock starts ticking at infection).

Total time of being sick is a real valued quantity aka it comes from an uncountable source (real line). I think you want the exponential distribution here, not the Poisson for ##X##.
- - - -
I also could not tell whether the rate of becoming infectious (characterized by ##\lambda##) and the rate of getting healthy (characterized by ##\lambda##) are independent or sequential processes? Could a person get healthy and never have the 'arrival' of the infectious state? (Or from a modeling standpoint, the infectious arrival happens after regaining health arrival and hence is moot.) I'm thinking the answer is yes, and this can be thought of as a case of Poisson splitting and combining if you wanted-- and again we're interested specifically in the time until next arrival. The interpretation here is needed for (c), and really (b)... Alternatively, these just happen in a fixed sequence.

Setting aside my interpretation questions, your part b is close, but not quite right... (why? start by looking up the formula of the CDF for an exponential distribution ). You should be comfortable using CDFs -- they are quite powerful.

I'll leave (c) open for now as the (in)dependence of infectiousness and regaining health arrivals is needed first.
 
  • #4
haruspex said:
I wonder if you have misread the question. X covers the time both to become infectious and then to recover. Each of those has an expected duration of 1/λ. You answer to a) would give E(X)=1/λ.

I agree, and of course ##X## does not have a Poisson distribution, or anything like it.
 
  • #5
Ray Vickson said:
I agree, and of course ##X## does not have a Poisson distribution, or anything like it.
Funnily enough, I'm not so sure now. I may have misread it.
The problem is the past tense, "took":
Mark53 said:
independent of the time it took for them to become infectious
That made me think this time starts on becoming infectious. But, of course, as @StoneTemplePython points out, they could recover without becoming infectious, so logically the two times should start on being infected. That makes a) and b) trivial but c) somewhat tricky.
 
  • #6
StoneTemplePython said:
I'm not totally happy with the way this problem is worded. Keep in mind that a Poisson is a discrete distribution (or counting process) characterized by exponentially distributed inter-arrival times. ##X##, here is the total amount of time a given person has the virus (clock starts ticking at infection).

Setting aside my interpretation questions, your part b is close, but not quite right... (why? start by looking up the formula of the CDF for an exponential distribution ). You should be comfortable using CDFs -- they are quite powerful.

I'll leave (c) open for now as the (in)dependence of infectiousness and regaining health arrivals is needed first.

for part b

P(x<1)=intergral from 0 to 1 of λe^(-λx)=1-e^(-λ)

would this be correct for part b now?

how would I get started on part c?
 
  • #7
Mark53 said:
for part b

P(x<1)=intergral from 0 to 1 of λe^(-λx)=1-e^(-λ)

would this be correct for part b now?

how would I get started on part c?
The first thing that has to be resolved is the interpretation of the question. Do you read it that the recovery time starts immediately on being infected (so recovery may occur without ever becoming infectious) or on becoming infectious?
To me, the first makes logical sense in the real world, but the wording of the question suggests the second.
 
  • #8
haruspex said:
The first thing that has to be resolved is the interpretation of the question. Do you read it that the recovery time starts immediately on being infected (so recovery may occur without ever becoming infectious) or on becoming infectious?
To me, the first makes logical sense in the real world, but the wording of the question suggests the second.
From the question it seems like the second one
 
  • #9
Mark53 said:
From the question it seems like the second one
Ok.
The next problem is units. You are given a numerical value for λ but not the units. If you assume the units are day-1 then your answer to b) is correct.
For c), you have two conditions to be met. What is the probability of each? Are they independent?
 
  • #10
haruspex said:
Ok.
The next problem is units. You are given a numerical value for λ but not the units. If you assume the units are day-1 then your answer to b) is correct.
For c), you have two conditions to be met. What is the probability of each? Are they independent?
the questions says that they are independent of each other

and the probability would be the same for each as in part b but just substitute λ=1 into them
 
  • #11
Mark53 said:
the questions says that they are independent of each other

and the probability would be the same for each as in part b but just substitute λ=1 into them
Right, so what is the answer to c)?
 
  • #12
haruspex said:
Right, so what is the answer to c)?
P(x<1)*P(x<1)
(1-e^(-1))*(1-e^(-1))
=1-2e^(-1)+e^(-2)
=0.3996
 
  • #13
Mark53 said:
P(x<1)*P(x<1)
(1-e^(-1))*(1-e^(-1))
=1-2e^(-1)+e^(-2)
=0.3996
Looks good to me.
 
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Likes Mark53
  • #14
haruspex said:
Looks good to me.
thanks for the help!
 
  • #15
So with the interpretation we're going for (Infected -> Infectious -> Cured, and the underlying time scale is days), I think the answer for (c) is correct.

There are some technical nits for (a) and (b) though.

Consider ##Y## which is an exponential random variable with parameter ##\lambda##. ##Y_1## is time from infected --> infectious. Then ##Y_2## is time from infectious --> healthy.

As I understand it total time sick is given by ##X## where ##X = Y_1 + Y_2##. That is, the total time sick is actually the convolution of two exponential r.v.'s -- aka a distribution that is Erlang of order 2 (which you can get from cleverly differentiating the Poisson distribution if you prefer that to convolutions). A quick sanity check tells us ##E[X] = E[Y_1] + E[Y_2] = \frac{2}{\lambda}##, which conforms with the fact that on average it takes ##\frac{1}{\lambda}## for someone to go infected --> infectious, then another ##\frac{1}{\lambda}## to go from infectious --> healthy

Mark53 said:
(b) What is the probability an infected individual has this virus for less than 1 day?

I take it this means a person is infected (but not yet infectious) and we are interested in the probability they will be totally cured a day later. Thus we are interested in the probability of being 2 arrivals (or more -- after 2 occur they are moot in our model) in a day, or alternatively -- we are interested in the complement of there being only 0 or 1 arrivals in a day -- note that this is the CDF of an Erlang of order 2.

If you haven't heard of an Erlang distribution, apologies -- they very natural come up when discussing Poisson processes and convolutions of exponentials.
 

Related to Probability virus question at different infection rates

1. What is the probability of getting infected with a virus at different infection rates?

The probability of getting infected with a virus at different infection rates depends on several factors such as the transmission rate of the virus, the susceptibility of the individual, and the number of people in the population who are infected. The higher the infection rate, the higher the probability of getting infected.

2. How does the infection rate affect the spread of a virus?

The infection rate directly affects the spread of a virus. A higher infection rate means that more people are getting infected, which leads to a faster spread of the virus. This is because infected individuals have a higher chance of transmitting the virus to others.

3. Can the infection rate of a virus change over time?

Yes, the infection rate of a virus can change over time. Factors such as the effectiveness of public health measures, the development of vaccines or treatments, and changes in human behavior can all affect the infection rate. Thus, it is important to monitor and track the infection rate of a virus to better understand its spread.

4. How does the probability of getting infected change with different levels of contact?

The probability of getting infected can change with different levels of contact depending on the mode of transmission of the virus. For example, if the virus is transmitted through close contact, such as respiratory droplets, the probability of getting infected is higher when in close proximity to an infected individual. However, if the virus is transmitted through surfaces, the probability of getting infected may be higher with frequent contact with contaminated surfaces.

5. How can we calculate the probability of getting infected with a virus at a specific infection rate?

The calculation of the probability of getting infected with a virus at a specific infection rate is complex and involves factors such as the transmission rate, the susceptibility of the individual, and the number of people in the population who are infected. Mathematical models and simulations are often used to estimate the probability of infection at different infection rates.

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