Finding the moment generating function of $min(Y,1)$
$begingroup$
Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.
By simply calculating the probability I managed to find the CDF of $X$ is:
$$F_X(t) = begin{cases}
1-e^{-t}, & 0 le t <1\
1, & t ge 1\
0, & text{otherwise}
end{cases}$$
Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).
I will appreciate some help.
probability moment-generating-functions exponential-distribution
edited Dec 10 '18 at 9:47
Saad
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
$endgroup$
add a comment |
$begingroup$
Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.
By simply calculating the probability I managed to find the CDF of $X$ is:
$$F_X(t) = begin{cases}
1-e^{-t}, & 0 le t <1\
1, & t ge 1\
0, & text{otherwise}
end{cases}$$
Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).
I will appreciate some help.
probability moment-generating-functions exponential-distribution
edited Dec 10 '18 at 9:47
Saad
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
$endgroup$
add a comment |
$begingroup$
Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.
By simply calculating the probability I managed to find the CDF of $X$ is:
$$F_X(t) = begin{cases}
1-e^{-t}, & 0 le t <1\
1, & t ge 1\
0, & text{otherwise}
end{cases}$$
Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).
I will appreciate some help.
probability moment-generating-functions exponential-distribution
edited Dec 10 '18 at 9:47
Saad
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
$endgroup$
Let $Ysimtext{Exp}(1)$ be a random variable. I denote the random variable $X$ as $X=min(Y,1)$. The task is to find the moment generating function of $X$.
By simply calculating the probability I managed to find the CDF of $X$ is:
$$F_X(t) = begin{cases}
1-e^{-t}, & 0 le t <1\
1, & t ge 1\
0, & text{otherwise}
end{cases}$$
Here I got stuck. Since $X$ is not a continuous variable, it does not have a PDF, and without it I do not know how to calculate the moment generating function. ($X$ is also not discrete).
I will appreciate some help.
probability moment-generating-functions exponential-distribution
probability moment-generating-functions exponential-distribution
edited Dec 10 '18 at 9:47
Saad
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
edited Dec 10 '18 at 9:47
Saad
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
edited Dec 10 '18 at 9:47
Saad
19.7k92352
edited Dec 10 '18 at 9:47
Saad
19.7k92352
edited Dec 10 '18 at 9:47
Saad
19.7k92352
19.7k92352
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
asked Dec 10 '18 at 0:52
Gabi GGabi G
39819
39819
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
The definition of the moment generating function does not require a PDF.
$$phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write
$$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
$$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The definition of the moment generating function does not require a PDF.
$$phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write
$$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
$$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
add a comment |
$begingroup$
The definition of the moment generating function does not require a PDF.
$$phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write
$$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
$$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
add a comment |
$begingroup$
The definition of the moment generating function does not require a PDF.
$$phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write
$$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
$$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
$endgroup$
The definition of the moment generating function does not require a PDF.
$$phi_X(t) := E[e^{tX}].$$
From here, it may be helpful to write
$$e^{tX} = e^{tX} mathbf{1}{Y > 1} + e^{tX} mathbf{1}{Y le 1}$$
and compute the expectation of the two terms separately.
Edit for more details:
$$E[e^{tX} mathbf{1}{Y > 1}] = E[e^{tY} mathbf{1}{Y > 1}] = int_1^infty e^{ty} f_Y(y) , dy = cdots.$$
$$E[e^{tX} mathbf{1}{Y le 1}] = e^t E[mathbf{1}{Y le 1}] = e^t P(Y le 1) = cdots.$$
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
edited Dec 10 '18 at 17:11
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
answered Dec 10 '18 at 1:00
angryavianangryavian
40.6k23380
40.6k23380
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
add a comment |
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Could you show me how?
$endgroup$
– Gabi G
Dec 10 '18 at 7:25
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
$begingroup$
Thank you! But I think the integral needs to be from 0 to 1
$endgroup$
– Gabi G
Dec 10 '18 at 11:59
add a comment |
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