Deriving mean through MGF for discrete uniform distribution
0
$begingroup$
The MGF of a discrete uniform distribution is given as
$$M_X(t) = frac{e^t(1-(e^t)^k)}{k(1-e^t)}$$
Looking for $E(X)$, I am computing $$M_X'(t) =frac{(k(e^t-1)-1)e^{(kt+t)}+e^t}{ k(e^t-1)^2}$$
which doesn't make sense cause denominator will become zero and mean should be $$E(X)=frac{k+1}{2}$$
How do I get the mean from the MGF? I can't seem to find out where I went wrong.
probability-theory generating-functions expected-value
edited Dec 2 '18 at 11:05
Did
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
$endgroup$
|
show 1 more comment
0
$begingroup$
The MGF of a discrete uniform distribution is given as
$$M_X(t) = frac{e^t(1-(e^t)^k)}{k(1-e^t)}$$
Looking for $E(X)$, I am computing $$M_X'(t) =frac{(k(e^t-1)-1)e^{(kt+t)}+e^t}{ k(e^t-1)^2}$$
which doesn't make sense cause denominator will become zero and mean should be $$E(X)=frac{k+1}{2}$$
How do I get the mean from the MGF? I can't seem to find out where I went wrong.
probability-theory generating-functions expected-value
edited Dec 2 '18 at 11:05
Did
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
$endgroup$
1
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
1
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29
|
show 1 more comment
0
0
0
$begingroup$
The MGF of a discrete uniform distribution is given as
$$M_X(t) = frac{e^t(1-(e^t)^k)}{k(1-e^t)}$$
Looking for $E(X)$, I am computing $$M_X'(t) =frac{(k(e^t-1)-1)e^{(kt+t)}+e^t}{ k(e^t-1)^2}$$
which doesn't make sense cause denominator will become zero and mean should be $$E(X)=frac{k+1}{2}$$
How do I get the mean from the MGF? I can't seem to find out where I went wrong.
probability-theory generating-functions expected-value
edited Dec 2 '18 at 11:05
Did
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
$endgroup$
The MGF of a discrete uniform distribution is given as
$$M_X(t) = frac{e^t(1-(e^t)^k)}{k(1-e^t)}$$
Looking for $E(X)$, I am computing $$M_X'(t) =frac{(k(e^t-1)-1)e^{(kt+t)}+e^t}{ k(e^t-1)^2}$$
which doesn't make sense cause denominator will become zero and mean should be $$E(X)=frac{k+1}{2}$$
How do I get the mean from the MGF? I can't seem to find out where I went wrong.
probability-theory generating-functions expected-value
probability-theory generating-functions expected-value
edited Dec 2 '18 at 11:05
Did
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
edited Dec 2 '18 at 11:05
Did
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
edited Dec 2 '18 at 11:05
Did
248k23224462
edited Dec 2 '18 at 11:05
Did
248k23224462
edited Dec 2 '18 at 11:05
Did
248k23224462
248k23224462
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
asked Dec 2 '18 at 10:59
Sumukh SaiSumukh Sai
206
206
1
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
1
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29
|
show 1 more comment
1
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
1
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29
1
1
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
1
1
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29
|
show 1 more comment
0
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1
$begingroup$
It very much makes sense because the numerator also goes to zero when $tto0$. Thus, simply use $E(X)=M_X(0)$ by computing $$M_X(0)=lim_{tto0}M_X(t)$$
$endgroup$
– Did
Dec 2 '18 at 11:03
$begingroup$
This is the first time I'm coming across the formula. Do you know where I could look it up
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:09
$begingroup$
Which formula? You invoke yourself the fact that $E(X)=M_X(0)$. The fact that $M_X(0)=limlimits_{tto0}M_X(t)$ is the continuity of $M_X$ at $0$, a most standard property, holding as soon as $E(X)$ is finite, and explained almost everywhere.
$endgroup$
– Did
Dec 2 '18 at 11:14
$begingroup$
I thought $E[X]= M_{X}'(0)$. This is the first time I came across $E[X] = M_{X}(0)$
$endgroup$
– Sumukh Sai
Dec 2 '18 at 11:17
1
$begingroup$
OK, obvious typo in my comments, please read $E(X)=M'_X(0)=limlimits_{tto0}M'_X(t)$.
$endgroup$
– Did
Dec 2 '18 at 11:29