Mixing
Expectation of a continuous random variable
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I am looking at a math problem that is asking me to determine $mathbbE[e^X]$ given that $X$ is a random variable, and the probability density function of $X$ is given by
$$ f(x) =begincases
1 & 0 leq x leq 1 \
0 & textotherwise.
endcases
$$
The solution from the answer key is below:
Let $Y = e^X.$ Then we wish to determine $mathbbE[Y]$. But first, we need to find the probability density function $f_y$ of $Y$. To do that, we begin with the probability cumulative distribution function $F_Y$ of $Y$:
$$
beginalign
F_Y = PY leq x \
= Pe^X leq x \
= PX leq log(x) \
= int_0^log(x) f(y) mathopdy \
= log(x).
endalign
$$
Differentiating, we obtain $f_Y = dfrac1x$. Then we have $$mathbbE[e^X] = mathbbE[Y] = int_-infty^infty xf_Y(x) mathopdx $$
$$= e - 1.$$
My question:
I don't understand why the integrand is $f(y)$ and not $f(x)$ when we're determining $F_Y$. I thought $PX leq log(x) = int_-infty^log(x) f_X(x)$, where $f_X$ is the probability density function of $X$.
probability
asked yesterday
Hat
745115
add a comment |Â
up vote
1
down vote
favorite
I am looking at a math problem that is asking me to determine $mathbbE[e^X]$ given that $X$ is a random variable, and the probability density function of $X$ is given by
$$ f(x) =begincases
1 & 0 leq x leq 1 \
0 & textotherwise.
endcases
$$
The solution from the answer key is below:
Let $Y = e^X.$ Then we wish to determine $mathbbE[Y]$. But first, we need to find the probability density function $f_y$ of $Y$. To do that, we begin with the probability cumulative distribution function $F_Y$ of $Y$:
$$
beginalign
F_Y = PY leq x \
= Pe^X leq x \
= PX leq log(x) \
= int_0^log(x) f(y) mathopdy \
= log(x).
endalign
$$
Differentiating, we obtain $f_Y = dfrac1x$. Then we have $$mathbbE[e^X] = mathbbE[Y] = int_-infty^infty xf_Y(x) mathopdx $$
$$= e - 1.$$
My question:
I don't understand why the integrand is $f(y)$ and not $f(x)$ when we're determining $F_Y$. I thought $PX leq log(x) = int_-infty^log(x) f_X(x)$, where $f_X$ is the probability density function of $X$.
probability
asked yesterday
Hat
745115
1
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
1
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
1
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am looking at a math problem that is asking me to determine $mathbbE[e^X]$ given that $X$ is a random variable, and the probability density function of $X$ is given by
$$ f(x) =begincases
1 & 0 leq x leq 1 \
0 & textotherwise.
endcases
$$
The solution from the answer key is below:
Let $Y = e^X.$ Then we wish to determine $mathbbE[Y]$. But first, we need to find the probability density function $f_y$ of $Y$. To do that, we begin with the probability cumulative distribution function $F_Y$ of $Y$:
$$
beginalign
F_Y = PY leq x \
= Pe^X leq x \
= PX leq log(x) \
= int_0^log(x) f(y) mathopdy \
= log(x).
endalign
$$
Differentiating, we obtain $f_Y = dfrac1x$. Then we have $$mathbbE[e^X] = mathbbE[Y] = int_-infty^infty xf_Y(x) mathopdx $$
$$= e - 1.$$
My question:
I don't understand why the integrand is $f(y)$ and not $f(x)$ when we're determining $F_Y$. I thought $PX leq log(x) = int_-infty^log(x) f_X(x)$, where $f_X$ is the probability density function of $X$.
probability
asked yesterday
Hat
745115
I am looking at a math problem that is asking me to determine $mathbbE[e^X]$ given that $X$ is a random variable, and the probability density function of $X$ is given by
$$ f(x) =begincases
1 & 0 leq x leq 1 \
0 & textotherwise.
endcases
$$
The solution from the answer key is below:
Let $Y = e^X.$ Then we wish to determine $mathbbE[Y]$. But first, we need to find the probability density function $f_y$ of $Y$. To do that, we begin with the probability cumulative distribution function $F_Y$ of $Y$:
$$
beginalign
F_Y = PY leq x \
= Pe^X leq x \
= PX leq log(x) \
= int_0^log(x) f(y) mathopdy \
= log(x).
endalign
$$
Differentiating, we obtain $f_Y = dfrac1x$. Then we have $$mathbbE[e^X] = mathbbE[Y] = int_-infty^infty xf_Y(x) mathopdx $$
$$= e - 1.$$
My question:
I don't understand why the integrand is $f(y)$ and not $f(x)$ when we're determining $F_Y$. I thought $PX leq log(x) = int_-infty^log(x) f_X(x)$, where $f_X$ is the probability density function of $X$.
probability
asked yesterday
Hat
745115
asked yesterday
Hat
745115
asked yesterday
Hat
745115
asked yesterday
Hat
745115
745115
1
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
1
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
1
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday
add a comment |Â
1
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
1
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
1
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday
1
1
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
1
1
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
1
1
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The solution is confusing you by using $y$ for two different things. In the integral, $y$ is just a dummy variable. They could just as easily have used $t$ instead of $y$.
answered yesterday
Brian Tung
25.1k32353
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The solution is confusing you by using $y$ for two different things. In the integral, $y$ is just a dummy variable. They could just as easily have used $t$ instead of $y$.
answered yesterday
Brian Tung
25.1k32353
add a comment |Â
up vote
3
down vote
accepted
The solution is confusing you by using $y$ for two different things. In the integral, $y$ is just a dummy variable. They could just as easily have used $t$ instead of $y$.
answered yesterday
Brian Tung
25.1k32353
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The solution is confusing you by using $y$ for two different things. In the integral, $y$ is just a dummy variable. They could just as easily have used $t$ instead of $y$.
answered yesterday
Brian Tung
25.1k32353
The solution is confusing you by using $y$ for two different things. In the integral, $y$ is just a dummy variable. They could just as easily have used $t$ instead of $y$.
answered yesterday
Brian Tung
25.1k32353
answered yesterday
Brian Tung
25.1k32353
answered yesterday
Brian Tung
25.1k32353
answered yesterday
Brian Tung
25.1k32353
25.1k32353
add a comment |Â
add a comment |Â
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1
Limit of the integral depends on $x$, so it is better to use $y$ as a dummy variable instead of $x$ itself to avoid confusion. And when one writes $f(y)$, it means the exact same pdf of $X$ as defined in the beginning.
– StubbornAtom
yesterday
1
Instead of finding the density of $Y$, it is simpler to say $E(e^X)=int_0^1 e^x f(x),dx$ using the Law of the unconscious statistician.
– StubbornAtom
yesterday
Your explanation makes sense. I did not know about that law. It looks much faster that way. Thank you!
– Hat
yesterday
1
@StubbornAtom: I have never heard that theorem given a name. About as good as any, I guess!
– Brian Tung
yesterday