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Given two exponential random variables $X_1$ and $X_2$ and their expected values, how do i approach to find the expected value of $Y = min(X_1, X_2)$
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Here's the problem statement :
``Let $X_1$ and $X_2$ be two exponentially distributed random variables with mean $0.5$ and $0.25$. What is the mean of $Y = min(X_1, X_2)$? ''
We know, the probability density function of exponential function is as follows :
$f(x) = lambda e^-lambda x$ ; $x geq 0$ and $0$ everywhere else.
Also, $E(x) = frac1lambda$ and hence $lambda_1 = frac10.5 = 2$ and $lambda_2 = frac10.25 = 4$
The official solution is as follows :
$$E(Y) = frac1lambda_1 + lambda_2 = frac12 + 4 = frac 16$$
Please help me understand the solution !
Thank you so much.
probability probability-theory probability-distributions
asked Jul 29 at 23:22
bluespring
85
add a comment |Â
up vote
1
down vote
favorite
Here's the problem statement :
``Let $X_1$ and $X_2$ be two exponentially distributed random variables with mean $0.5$ and $0.25$. What is the mean of $Y = min(X_1, X_2)$? ''
We know, the probability density function of exponential function is as follows :
$f(x) = lambda e^-lambda x$ ; $x geq 0$ and $0$ everywhere else.
Also, $E(x) = frac1lambda$ and hence $lambda_1 = frac10.5 = 2$ and $lambda_2 = frac10.25 = 4$
The official solution is as follows :
$$E(Y) = frac1lambda_1 + lambda_2 = frac12 + 4 = frac 16$$
Please help me understand the solution !
Thank you so much.
probability probability-theory probability-distributions
asked Jul 29 at 23:22
bluespring
85
@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here's the problem statement :
``Let $X_1$ and $X_2$ be two exponentially distributed random variables with mean $0.5$ and $0.25$. What is the mean of $Y = min(X_1, X_2)$? ''
We know, the probability density function of exponential function is as follows :
$f(x) = lambda e^-lambda x$ ; $x geq 0$ and $0$ everywhere else.
Also, $E(x) = frac1lambda$ and hence $lambda_1 = frac10.5 = 2$ and $lambda_2 = frac10.25 = 4$
The official solution is as follows :
$$E(Y) = frac1lambda_1 + lambda_2 = frac12 + 4 = frac 16$$
Please help me understand the solution !
Thank you so much.
probability probability-theory probability-distributions
asked Jul 29 at 23:22
bluespring
85
Here's the problem statement :
``Let $X_1$ and $X_2$ be two exponentially distributed random variables with mean $0.5$ and $0.25$. What is the mean of $Y = min(X_1, X_2)$? ''
We know, the probability density function of exponential function is as follows :
$f(x) = lambda e^-lambda x$ ; $x geq 0$ and $0$ everywhere else.
Also, $E(x) = frac1lambda$ and hence $lambda_1 = frac10.5 = 2$ and $lambda_2 = frac10.25 = 4$
The official solution is as follows :
$$E(Y) = frac1lambda_1 + lambda_2 = frac12 + 4 = frac 16$$
Please help me understand the solution !
Thank you so much.
probability probability-theory probability-distributions
asked Jul 29 at 23:22
bluespring
85
edited Jul 30 at 23:49
asked Jul 29 at 23:22
bluespring
85
asked Jul 29 at 23:22
bluespring
85
asked Jul 29 at 23:22
bluespring
85
85
@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29
add a comment |Â
@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29
@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$$E[Y] = int_0^inftyy,f_Y(y),dy$$
$$f_Y(y)=fracdF_Y(y)dy$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=textPleft[min(X_1, X_2)leq yright] \= 1- textPleft[min(X_1, X_2) > yright] \= 1-textPleft[X_1 > yright]textPleft[X_2 > yright]\=1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right]\=1-expleft(-y[lambda_1+lambda_2]right)$$
So
$$f_Y(y) = (lambda_1+lambda_2)expleft(-y[lambda_1+lambda_2]right)$$
Substitute in the first integration for the expectation and do the integration.
answered Jul 30 at 1:07
BlackMath
877
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
add a comment |Â
up vote
1
down vote
$f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $PY>t=PX_1>tPX_2>t=int_t^infty (0.5) e^-(0.5)s ds int_t^infty (0.25) e^-(0.25)s ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation.
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
add a comment |Â
up vote
1
down vote
If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $mathsf P(X_1leqslant X_2)=int_0^inftyint_0^t 8 e^- 2s-4tmathsf d smathsf d t=tfrac 13$
answered Jul 29 at 23:51
Graham Kemp
80k43275
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$E[Y] = int_0^inftyy,f_Y(y),dy$$
$$f_Y(y)=fracdF_Y(y)dy$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=textPleft[min(X_1, X_2)leq yright] \= 1- textPleft[min(X_1, X_2) > yright] \= 1-textPleft[X_1 > yright]textPleft[X_2 > yright]\=1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right]\=1-expleft(-y[lambda_1+lambda_2]right)$$
So
$$f_Y(y) = (lambda_1+lambda_2)expleft(-y[lambda_1+lambda_2]right)$$
Substitute in the first integration for the expectation and do the integration.
answered Jul 30 at 1:07
BlackMath
877
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
add a comment |Â
up vote
1
down vote
accepted
$$E[Y] = int_0^inftyy,f_Y(y),dy$$
$$f_Y(y)=fracdF_Y(y)dy$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=textPleft[min(X_1, X_2)leq yright] \= 1- textPleft[min(X_1, X_2) > yright] \= 1-textPleft[X_1 > yright]textPleft[X_2 > yright]\=1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right]\=1-expleft(-y[lambda_1+lambda_2]right)$$
So
$$f_Y(y) = (lambda_1+lambda_2)expleft(-y[lambda_1+lambda_2]right)$$
Substitute in the first integration for the expectation and do the integration.
answered Jul 30 at 1:07
BlackMath
877
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$E[Y] = int_0^inftyy,f_Y(y),dy$$
$$f_Y(y)=fracdF_Y(y)dy$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=textPleft[min(X_1, X_2)leq yright] \= 1- textPleft[min(X_1, X_2) > yright] \= 1-textPleft[X_1 > yright]textPleft[X_2 > yright]\=1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right]\=1-expleft(-y[lambda_1+lambda_2]right)$$
So
$$f_Y(y) = (lambda_1+lambda_2)expleft(-y[lambda_1+lambda_2]right)$$
Substitute in the first integration for the expectation and do the integration.
answered Jul 30 at 1:07
BlackMath
877
$$E[Y] = int_0^inftyy,f_Y(y),dy$$
$$f_Y(y)=fracdF_Y(y)dy$$
Assuming $X_1$ and $X_2$ are independent, we have
$$F_Y(y)=textPleft[min(X_1, X_2)leq yright] \= 1- textPleft[min(X_1, X_2) > yright] \= 1-textPleft[X_1 > yright]textPleft[X_2 > yright]\=1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right]\=1-expleft(-y[lambda_1+lambda_2]right)$$
So
$$f_Y(y) = (lambda_1+lambda_2)expleft(-y[lambda_1+lambda_2]right)$$
Substitute in the first integration for the expectation and do the integration.
answered Jul 30 at 1:07
BlackMath
877
edited Jul 30 at 2:06
answered Jul 30 at 1:07
BlackMath
877
answered Jul 30 at 1:07
BlackMath
877
answered Jul 30 at 1:07
BlackMath
877
877
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
add a comment |Â
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring $$1-left[1-textPleft[X_1leq yright]right]left[1-textPleft[X_2leq yright]right] = 1-left[1-left1-exp(-lambda_1,y)rightright]left[1-left1-exp(-lambda_2,y)rightright]=1-expleft(-y[lambda_1+lambda_2]right)$$
– BlackMath
Jul 30 at 2:04
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
@bluespring If there is a simpler way, I would like to hear it, too. Simpler is better. Is there something that you find difficult to understand in my answer? I will be glad to explain more.
– BlackMath
Jul 30 at 3:55
add a comment |Â
up vote
1
down vote
$f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $PY>t=PX_1>tPX_2>t=int_t^infty (0.5) e^-(0.5)s ds int_t^infty (0.25) e^-(0.25)s ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation.
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
add a comment |Â
up vote
1
down vote
$f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $PY>t=PX_1>tPX_2>t=int_t^infty (0.5) e^-(0.5)s ds int_t^infty (0.25) e^-(0.25)s ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation.
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $PY>t=PX_1>tPX_2>t=int_t^infty (0.5) e^-(0.5)s ds int_t^infty (0.25) e^-(0.25)s ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation.
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
$f(X_2)<f(X_1)$ is wrong. The correct method is as follows: $PY>t=PX_1>tPX_2>t=int_t^infty (0.5) e^-(0.5)s ds int_t^infty (0.25) e^-(0.25)s ds$. Compute the integrals and differentiate to get the density of $Y$. Then compute its expectation.
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
edited Jul 29 at 23:38
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
answered Jul 29 at 23:36
Kavi Rama Murthy
19.7k2829
19.7k2829
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
add a comment |Â
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
@ZacharySelk Thanks for the correction.
– Kavi Rama Murthy
Jul 29 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
Thank you sir for your kind help. I realize my error now.
– bluespring
Jul 30 at 23:39
add a comment |Â
up vote
1
down vote
If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $mathsf P(X_1leqslant X_2)=int_0^inftyint_0^t 8 e^- 2s-4tmathsf d smathsf d t=tfrac 13$
answered Jul 29 at 23:51
Graham Kemp
80k43275
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
add a comment |Â
up vote
1
down vote
If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $mathsf P(X_1leqslant X_2)=int_0^inftyint_0^t 8 e^- 2s-4tmathsf d smathsf d t=tfrac 13$
answered Jul 29 at 23:51
Graham Kemp
80k43275
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $mathsf P(X_1leqslant X_2)=int_0^inftyint_0^t 8 e^- 2s-4tmathsf d smathsf d t=tfrac 13$
answered Jul 29 at 23:51
Graham Kemp
80k43275
If $X_1,X_2$ are independent, it is explicitly possible that the event of $X_1<X_2$ may occur.
The fact that the probability denisity (or cumulative distribution) function of $X_2$ is greater than that of $X_1$ at every point does not mean $X_2$ will always have the lesser value; rather it suggests that it is more likely to.
Indeed $mathsf P(X_1leqslant X_2)=int_0^inftyint_0^t 8 e^- 2s-4tmathsf d smathsf d t=tfrac 13$
answered Jul 29 at 23:51
Graham Kemp
80k43275
edited Jul 29 at 23:55
answered Jul 29 at 23:51
Graham Kemp
80k43275
answered Jul 29 at 23:51
Graham Kemp
80k43275
answered Jul 29 at 23:51
Graham Kemp
80k43275
80k43275
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
add a comment |Â
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
This is some seriously advanced maths. I'm not able to follow a lot here. I'm from the engineering background, so we haven't learnt all this.
– bluespring
Jul 29 at 23:53
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
but thank you really for your response, its very much appreciated !
– bluespring
Jul 30 at 0:00
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@ZacharySelk its the pdf of exponential distribution $f(x) = lambda e^-lambda x ; x geq 0$ and $0$ everywhere else
– bluespring
Jul 29 at 23:35
The official solution looks sensible. One illustration is that you arrive at a bus stop served by two routes, one with a service of $2$ buses an hour and the other with a service of $4$ buses an hour, the two routes being independent Poisson processes, so the expected time until a bus on the first route is $frac12$ hour and the expected time until a bus on the second route is $frac14$ hour. You are only going a short distance where the routes coincide, so do not care which arrives first. For you the rate is $6$ buses an hour and the expected time until the first bus is $frac16$ hour
– Henry
Jul 30 at 0:14
@Henry Perhaps you're right ! i was however expecting a little conventional approach. your method reminds me of queuing theory, where the arrival distribution is poisson and the service distribution is exponential. However, you've assumed arrival to be exponential here. Again, i'm not very good at maths, so pardon me.
– bluespring
Jul 30 at 0:29