# Stochastic Processes

• Dec 8th 2009, 03:10 AM
asingh88
Exponential Distribution
Suppose three particles are particle projected, P, Q and R. Each particle is projected once. The projection is modelled as an exponentially distributed random variable. Assuming that the average projections for P, Q and R are 50m, 60m and 70m respectively, find:

(a) Probability of a particle being projected 150m or further.
(b) Probability that P will go further than R.
• Dec 8th 2009, 07:36 AM
novice
Quote:

Originally Posted by asingh88
Suppose three particles are particle projected, P, Q and R. Each particle is projected once. The projection is modelled as an exponentially distributed random variable. Assuming that the average projections for P, Q and R are 50m, 60m and 70m respectively, find:

(a) Probability of a particle being projected 150m or further.
(b) Probability that P will go further than R.

$\lambda_P=50$
$\lambda_Q=60$
$\lambda_R=70$

Probability of a particle being projected 150m or further:

$P(X_P\geq 150)=e^{-x/\lambda_P}$
$P(X_Q\geq 150)=e^{-x/\lambda_Q}$
$P(X_R\geq 150)=e^{-x/\lambda_R}$

Probability that P will go further than R:

$e^{-x/\lambda_R} \geq e^{-x/\lambda_P}\Longrightarrow P(X_P>X_R)=\emptyset$
• Dec 8th 2009, 08:12 AM
qmech
If we call
$\lambda_P=50$

then the exponential probability distribution
$
P_P(x)=\frac{e^{-x/\lambda_P}}{\lambda_P}
$

(which is normalized so the total probability is 1, i.e., the integral from 0 to infinity of the probability distribution is 1.)

The probability of an x >= 150 should be:
$
P_P(x \geq 150)=\int_{150}^{\infty} \frac{e^{-x/\lambda_P}}{\lambda_P} dx
$

The probability that one particle goes farther than another should be
$
P(X_P \geq X_R)=
\int_{0}^{\infty} P_P(y)dy
\int_{0}^{y} P_R(x)dx
$

in other words choose some value for the variable with the P distribution (i.e., y). Find the probability for that y. Now multiply that by the probability that the other variable has a value less than that. Now integrate over all possible values of y. Note $X_P$ is the variable related to $\lambda_P$, and $X_R$ is the variable related to $\lambda_R$
• Dec 8th 2009, 10:06 AM
asingh88
C) probability that the winner of the particle competition will beat the current particle world record of 196m
• Dec 8th 2009, 04:46 PM
Focus
Quote:

Originally Posted by asingh88
C) probability that the winner of the particle competition will beat the current particle world record of 196m

I don't mean to be rude here, but you need to show a bit of effort in this thread. You had two people respond to you which you haven't acknowledged and it seems to me that you just want answers to some homework.
• Dec 9th 2009, 10:22 AM
asingh88
its not like you answered the question,fair enough if you answered my question in the first place i would know where your coming from,the whole point of this forum is to help people not ridicule people that are stuck
• Dec 9th 2009, 11:33 AM
Quote:

Originally Posted by asingh88
Suppose three particles are particle projected, P, Q and R. Each particle is projected once. The projection is modelled as an exponentially distributed random variable. Assuming that the average projections for P, Q and R are 50m, 60m and 70m respectively, find:

(a) Probability of a particle being projected 150m or further.
(b) Probability that P will go further than R.

can you do this?:

a)

memoryless property

lamda = (1/50)+(1/60)+(1/70)

exp(-lambda * 150)

b)

lambda = 50

integrate with limits infinity and 70 1/lambda * exp(-x/lamda)

??
• Dec 10th 2009, 06:56 AM
Focus
Quote:

Originally Posted by asingh88
its not like you answered the question,fair enough if you answered my question in the first place i would know where your coming from,the whole point of this forum is to help people not ridicule people that are stuck

I am not ridiculing you, I am just pointing out the fact that it seems you have not attempted the questions. If you have post it here, even if it is wrong so that people at least can tell you how to correct it or go about it. If you have no idea where to start, say so. Like you said, the point of these forums are to help people and not to do their homework for them.

As for a hint on part c), you don't need to worry about the winner, just that one particle beats the world record. When you are working with the maximum, it is often better to calculate $\mathbb{P}(X_1 \vee X_2 \vee X_3 \leq a)=\mathbb{P}(X_1 \leq a , X_2 \leq a , X_3 \leq a)$.