Consider that an individual whose level of exposure to a certain pathogen is rx € [0, 0) will contract the disease caused by this pathogen with probability P(x) E (0, 1). Suppose that the exposure level of a randomly chosen member of the population is given by a non-negative continuous random variable X with probability density function fx (i.e., fx(x) > 0, for æ > 0, and fx(x) = 0, otherwise). Find the conditional probability density of the exposure level of that (random) member given that the individual has the disease.
Consider that an individual whose level of exposure to a certain pathogen is rx € [0, 0) will contract the disease caused by this pathogen with probability P(x) E (0, 1). Suppose that the exposure level of a randomly chosen member of the population is given by a non-negative continuous random variable X with probability density function fx (i.e., fx(x) > 0, for æ > 0, and fx(x) = 0, otherwise). Find the conditional probability density of the exposure level of that (random) member given that the individual has the disease.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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This is a question about Applied Stochastic Models.
Question and definition used are attached.
![Definition 21 (Continuous case-II). Let X be a (continuous or discrete) real random variable and let
Y be a continuous random variable. Suppose that the joint density fx,y of X and Y satisfies the properties
(a) and (b). Consider y e R such that P(y < Y < y + ɛ) > 0, for all e > 0, and fy(y) > 0, where fy is the
density function of Y. Then, the conditional distribution function of X given that Y = y is defined by
Ez«s fx.y(z.y)1 =s3| if X discrete,
Fx]Y=y(x) = P(X < x[Y = y) = {
for xe R.
fy(y)
if X continuous,
Furthermore, if X is a continuous random variable and Fx¡y=y(x) is differentiable at x e R, we can take
fxyx=p(x) = Fx[Y=p(x).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc001743-f440-47c5-93da-dbaba94275c9%2Fbb022e90-c163-477e-862a-2a0d75918a83%2F5wh5d1_processed.png&w=3840&q=75)
Transcribed Image Text:Definition 21 (Continuous case-II). Let X be a (continuous or discrete) real random variable and let
Y be a continuous random variable. Suppose that the joint density fx,y of X and Y satisfies the properties
(a) and (b). Consider y e R such that P(y < Y < y + ɛ) > 0, for all e > 0, and fy(y) > 0, where fy is the
density function of Y. Then, the conditional distribution function of X given that Y = y is defined by
Ez«s fx.y(z.y)1 =s3| if X discrete,
Fx]Y=y(x) = P(X < x[Y = y) = {
for xe R.
fy(y)
if X continuous,
Furthermore, if X is a continuous random variable and Fx¡y=y(x) is differentiable at x e R, we can take
fxyx=p(x) = Fx[Y=p(x).
Transcribed Image Text:Consider that an individual whose level of exposure to a certain pathogen is z € [0, 0)
will contract the disease caused by this pathogen with probability P(x) E (0, 1). Suppose
that the exposure level of a randomly chosen member of the population is given by a
non-negative continuous random variable X with probability density function fx (i.e.,
fx(x) > 0, for x > 0, and fx(x) = 0, otherwise). Find the conditional probability
density of the exposure level of that (random) member given that the individual has the
disease.
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