Mixing
Definite integral $int_0^infty e^{-y}e^{-xy}y^2dy$
$begingroup$
I successfully solved the following integral of $int _0^infty e^{-y}e^{-xy}y^2dy$ by using partial integration two times, which was pretty hell long, and since solving this integral was not the main issue in the exercise I was doing, I wondered if there's an easier way to integrate it. Is there?
Thanks!
(btw my result was $frac{2}{(x+1)^3}$)
P.S I am not familiar with gamma functions or anything beyond "common B.SC calculus".
calculus integration
edited Jan 23 at 21:32
J.G.
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
$endgroup$
add a comment |
$begingroup$
I successfully solved the following integral of $int _0^infty e^{-y}e^{-xy}y^2dy$ by using partial integration two times, which was pretty hell long, and since solving this integral was not the main issue in the exercise I was doing, I wondered if there's an easier way to integrate it. Is there?
Thanks!
(btw my result was $frac{2}{(x+1)^3}$)
P.S I am not familiar with gamma functions or anything beyond "common B.SC calculus".
calculus integration
edited Jan 23 at 21:32
J.G.
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
$endgroup$
$begingroup$
This integral looks very definite...
$endgroup$
– user
Jan 23 at 21:26
$begingroup$
@user I edited it to include limits to match the stated result. I'll fix the title.
$endgroup$
– J.G.
Jan 23 at 21:31
add a comment |
$begingroup$
I successfully solved the following integral of $int _0^infty e^{-y}e^{-xy}y^2dy$ by using partial integration two times, which was pretty hell long, and since solving this integral was not the main issue in the exercise I was doing, I wondered if there's an easier way to integrate it. Is there?
Thanks!
(btw my result was $frac{2}{(x+1)^3}$)
P.S I am not familiar with gamma functions or anything beyond "common B.SC calculus".
calculus integration
edited Jan 23 at 21:32
J.G.
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
$endgroup$
I successfully solved the following integral of $int _0^infty e^{-y}e^{-xy}y^2dy$ by using partial integration two times, which was pretty hell long, and since solving this integral was not the main issue in the exercise I was doing, I wondered if there's an easier way to integrate it. Is there?
Thanks!
(btw my result was $frac{2}{(x+1)^3}$)
P.S I am not familiar with gamma functions or anything beyond "common B.SC calculus".
calculus integration
calculus integration
edited Jan 23 at 21:32
J.G.
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
edited Jan 23 at 21:32
J.G.
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
edited Jan 23 at 21:32
J.G.
30.1k23048
edited Jan 23 at 21:32
J.G.
30.1k23048
edited Jan 23 at 21:32
J.G.
30.1k23048
30.1k23048
asked Jan 23 at 21:08
superuser123superuser123
48628
asked Jan 23 at 21:08
superuser123superuser123
48628
asked Jan 23 at 21:08
superuser123superuser123
48628
48628
$begingroup$
This integral looks very definite...
$endgroup$
– user
Jan 23 at 21:26
$begingroup$
@user I edited it to include limits to match the stated result. I'll fix the title.
$endgroup$
– J.G.
Jan 23 at 21:31
add a comment |
$begingroup$
This integral looks very definite...
$endgroup$
– user
Jan 23 at 21:26
$begingroup$
@user I edited it to include limits to match the stated result. I'll fix the title.
$endgroup$
– J.G.
Jan 23 at 21:31
$begingroup$
This integral looks very definite...
$endgroup$
– user
Jan 23 at 21:26
$begingroup$
This integral looks very definite...
$endgroup$
– user
Jan 23 at 21:26
$begingroup$
@user I edited it to include limits to match the stated result. I'll fix the title.
$endgroup$
– J.G.
Jan 23 at 21:31
$begingroup$
@user I edited it to include limits to match the stated result. I'll fix the title.
$endgroup$
– J.G.
Jan 23 at 21:31
add a comment |
5 Answers
5
active
oldest
votes
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To prove by induction that $int_0^infty y^n e^{-zy}dy=frac{n!}{y^{n+1}}$ (for $Re z>0$), note that $$int_0^infty e^{-zy}dy=frac{1}{z},,int_0^infty y^{n+1} e^{-zy}dy=-partial_zint_0^infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
$endgroup$
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
add a comment |
$begingroup$
begin{align}
int_0^infty e^{-(1+x)y}y^2dy &= frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}e^{-(1+x)y}bigg|_0^infty
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}
\
&= frac{2}{(1+x)^3}
end{align}
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
$endgroup$
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
1
$begingroup$
First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
$endgroup$
– zahbaz
Jan 24 at 8:55
$begingroup$
Why does it bring down the factors of y? I just don't get it :(
$endgroup$
– superuser123
Jan 24 at 13:01
$begingroup$
Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
$endgroup$
– zahbaz
Jan 24 at 20:43
add a comment |
$begingroup$
Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf
$$f(y) = lambda e^{-lambda y}, quad lambda,y>0,$$
we have
$$E[Y] = frac{1}{lambda},$$
$$var(Y) = frac{1}{lambda^2} implies E[Y^2] = frac{2}{lambda^2}.$$
That is,
$$lambdaint_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^2} implies int_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^3}.$$
Can you guess $lambda$ in your problem?
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
$endgroup$
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How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
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I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
add a comment |
$begingroup$
$I(x) = int_0^{infty} e^{-y}e^{-xy}y^2 dy$
Integration under the integral
$I(x) = frac {d}{dx} int_0^{infty} e^{-y}y^2(int e^{-xy} dx) dy = frac {d}{dx}int_0^{infty} e^{-y(1+x)}y dy$
And one more time.
$I(x) = frac {d^2}{dx^2}int_0^{infty} e^{-y(1+x)} dy = frac {d^2}{dx^2}frac {1}{1+x} = frac {1}{2(1+x)^3}$
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
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add a comment |
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Try separating the integral like this
$$int _0^infty e^{-y(x+1)}y^2dy=int _0^1 e^{-y(x+1)}y^2dy+int _1^infty e^{-y(x+1)}y^2dy$$
you can see that the first integral is a proper one hence the integral converges.
For the second one try taking the cases when $x+1 lt0 $, $x+1 =0$ and $x+1 gt0$
for $x+1 gt0$ from $$lim_{yto infty}frac{e^{-y(x+1)}y^2}{frac{1}{y2}}=0$$ this implies that $$int _1^infty e^{-y(x+1)}y^2dyleint _1^infty frac{1}{y^2}dy$$
the last integral converges, from the comparison test so will the integral $int _1^infty e^{-y(x+1)}y^2dy$
Can you try the other cases?
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To prove by induction that $int_0^infty y^n e^{-zy}dy=frac{n!}{y^{n+1}}$ (for $Re z>0$), note that $$int_0^infty e^{-zy}dy=frac{1}{z},,int_0^infty y^{n+1} e^{-zy}dy=-partial_zint_0^infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
$endgroup$
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
add a comment |
$begingroup$
To prove by induction that $int_0^infty y^n e^{-zy}dy=frac{n!}{y^{n+1}}$ (for $Re z>0$), note that $$int_0^infty e^{-zy}dy=frac{1}{z},,int_0^infty y^{n+1} e^{-zy}dy=-partial_zint_0^infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
$endgroup$
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
add a comment |
$begingroup$
To prove by induction that $int_0^infty y^n e^{-zy}dy=frac{n!}{y^{n+1}}$ (for $Re z>0$), note that $$int_0^infty e^{-zy}dy=frac{1}{z},,int_0^infty y^{n+1} e^{-zy}dy=-partial_zint_0^infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
$endgroup$
To prove by induction that $int_0^infty y^n e^{-zy}dy=frac{n!}{y^{n+1}}$ (for $Re z>0$), note that $$int_0^infty e^{-zy}dy=frac{1}{z},,int_0^infty y^{n+1} e^{-zy}dy=-partial_zint_0^infty y^{n+1} e^{-zy}dy.$$For your stated problem, you only need to differentiate the base case twice; the induction is in case you want to generalise it.
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
answered Jan 23 at 21:14
J.G.J.G.
30.1k23048
30.1k23048
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
add a comment |
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
$begingroup$
Well, that's interesting. Is this some kind of a known theorem?
$endgroup$
– superuser123
Jan 23 at 21:18
1
1
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
$begingroup$
@superuser123 en.wikipedia.org/wiki/Leibniz_integral_rule
$endgroup$
– J.G.
Jan 23 at 21:20
add a comment |
$begingroup$
begin{align}
int_0^infty e^{-(1+x)y}y^2dy &= frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}e^{-(1+x)y}bigg|_0^infty
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}
\
&= frac{2}{(1+x)^3}
end{align}
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
$endgroup$
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
1
$begingroup$
First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
$endgroup$
– zahbaz
Jan 24 at 8:55
$begingroup$
Why does it bring down the factors of y? I just don't get it :(
$endgroup$
– superuser123
Jan 24 at 13:01
$begingroup$
Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
$endgroup$
– zahbaz
Jan 24 at 20:43
add a comment |
$begingroup$
begin{align}
int_0^infty e^{-(1+x)y}y^2dy &= frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}e^{-(1+x)y}bigg|_0^infty
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}
\
&= frac{2}{(1+x)^3}
end{align}
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
$endgroup$
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
1
$begingroup$
First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
$endgroup$
– zahbaz
Jan 24 at 8:55
$begingroup$
Why does it bring down the factors of y? I just don't get it :(
$endgroup$
– superuser123
Jan 24 at 13:01
$begingroup$
Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
$endgroup$
– zahbaz
Jan 24 at 20:43
add a comment |
$begingroup$
begin{align}
int_0^infty e^{-(1+x)y}y^2dy &= frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}e^{-(1+x)y}bigg|_0^infty
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}
\
&= frac{2}{(1+x)^3}
end{align}
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
$endgroup$
begin{align}
int_0^infty e^{-(1+x)y}y^2dy &= frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}e^{-(1+x)y}bigg|_0^infty
\
&= frac{d^2}{dx^2}frac{1}{(1+x)}
\
&= frac{2}{(1+x)^3}
end{align}
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
answered Jan 23 at 21:29
zahbazzahbaz
8,43921937
8,43921937
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
1
$begingroup$
First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
$endgroup$
– zahbaz
Jan 24 at 8:55
$begingroup$
Why does it bring down the factors of y? I just don't get it :(
$endgroup$
– superuser123
Jan 24 at 13:01
$begingroup$
Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
$endgroup$
– zahbaz
Jan 24 at 20:43
add a comment |
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
1
$begingroup$
First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
$endgroup$
– zahbaz
Jan 24 at 8:55
$begingroup$
Why does it bring down the factors of y? I just don't get it :(
$endgroup$
– superuser123
Jan 24 at 13:01
$begingroup$
Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
$endgroup$
– zahbaz
Jan 24 at 20:43
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
$endgroup$
– superuser123
Jan 24 at 7:40
$begingroup$
Thanks! can you please explain the first step? $int_0^infty e^{-(1+x)y}y^2dy =frac{d^2}{dx^2}int_0^infty e^{-(1+x)y}dy $
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– superuser123
Jan 24 at 7:40
1
1
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First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
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– zahbaz
Jan 24 at 8:55
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First, notice that $(1+x)$ is a constant with respect to the integral. Then notice that taking derivatives of the exponential with respect to $(1+x)$ or equivalently $x$ in this case will bring down factors of $y$ $$frac{d^2}{dx^2}e^{-(1+x)y} = e^{-(1+x)y} y^2$$ So you can say the integral is equal to $int_0^infty frac{d^2}{dx^2}e^{-(1+x)y}dy$. Roughly speaking then, since the integral doesn't depend on $x$, we can then interchange the order of integration and differentiation to get $$frac{d^2}{dx^2} int_0^infty e^{-(1+x)y}dy$$ This method is often referring to as Feynman's trick.
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– zahbaz
Jan 24 at 8:55
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Why does it bring down the factors of y? I just don't get it :(
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– superuser123
Jan 24 at 13:01
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Why does it bring down the factors of y? I just don't get it :(
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– superuser123
Jan 24 at 13:01
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Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
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– zahbaz
Jan 24 at 20:43
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Nothing fancy is going on. We're exploiting the properties of a derivative. Recall Calc I. The derivative of an exponential function is equal to the exponential times the derivative of its argument (i.e. chain rule applies). Consider these examples: $$frac{d}{dx} e^{ax} = a e^{ax}$$ $$frac{d^2}{dx^2} e^{ax} = a frac{d}{dx} e^{ax} = a^2 e^{ax}$$
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– zahbaz
Jan 24 at 20:43
add a comment |
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Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf
$$f(y) = lambda e^{-lambda y}, quad lambda,y>0,$$
we have
$$E[Y] = frac{1}{lambda},$$
$$var(Y) = frac{1}{lambda^2} implies E[Y^2] = frac{2}{lambda^2}.$$
That is,
$$lambdaint_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^2} implies int_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^3}.$$
Can you guess $lambda$ in your problem?
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
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How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
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– J.G.
Jan 23 at 21:21
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I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
add a comment |
$begingroup$
Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf
$$f(y) = lambda e^{-lambda y}, quad lambda,y>0,$$
we have
$$E[Y] = frac{1}{lambda},$$
$$var(Y) = frac{1}{lambda^2} implies E[Y^2] = frac{2}{lambda^2}.$$
That is,
$$lambdaint_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^2} implies int_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^3}.$$
Can you guess $lambda$ in your problem?
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
$endgroup$
$begingroup$
How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
$begingroup$
I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
add a comment |
$begingroup$
Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf
$$f(y) = lambda e^{-lambda y}, quad lambda,y>0,$$
we have
$$E[Y] = frac{1}{lambda},$$
$$var(Y) = frac{1}{lambda^2} implies E[Y^2] = frac{2}{lambda^2}.$$
That is,
$$lambdaint_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^2} implies int_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^3}.$$
Can you guess $lambda$ in your problem?
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
$endgroup$
Since your questions are related to random variables, this answer uses an exponential RV to compute the integral. Note that for an exponential RV, $Y$, with pdf
$$f(y) = lambda e^{-lambda y}, quad lambda,y>0,$$
we have
$$E[Y] = frac{1}{lambda},$$
$$var(Y) = frac{1}{lambda^2} implies E[Y^2] = frac{2}{lambda^2}.$$
That is,
$$lambdaint_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^2} implies int_0^{infty} y^2 e^{-lambda y} dy = frac{2}{lambda^3}.$$
Can you guess $lambda$ in your problem?
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
edited Jan 23 at 21:31
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
answered Jan 23 at 21:20
Math LoverMath Lover
14.1k31437
14.1k31437
$begingroup$
How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
$begingroup$
I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
add a comment |
$begingroup$
How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
$begingroup$
I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
$begingroup$
How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
$begingroup$
How would you prove $E[Y^2]=2lambda^{-2}$ without evaluating the integral? (I'm guessing you'd use a moment-generating or characteristic function, but it's worth explaining in your answer.)
$endgroup$
– J.G.
Jan 23 at 21:21
$begingroup$
I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
$begingroup$
I am presuming that OP is aware of this relationship. I agree that calculating the variance itself requires evaluation of an integral. One approach is to use MGF to get $E[Y^n] = frac{n!}{lambda^n}$ which is quite easy to evaluate.
$endgroup$
– Math Lover
Jan 23 at 21:26
add a comment |
$begingroup$
$I(x) = int_0^{infty} e^{-y}e^{-xy}y^2 dy$
Integration under the integral
$I(x) = frac {d}{dx} int_0^{infty} e^{-y}y^2(int e^{-xy} dx) dy = frac {d}{dx}int_0^{infty} e^{-y(1+x)}y dy$
And one more time.
$I(x) = frac {d^2}{dx^2}int_0^{infty} e^{-y(1+x)} dy = frac {d^2}{dx^2}frac {1}{1+x} = frac {1}{2(1+x)^3}$
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$I(x) = int_0^{infty} e^{-y}e^{-xy}y^2 dy$
Integration under the integral
$I(x) = frac {d}{dx} int_0^{infty} e^{-y}y^2(int e^{-xy} dx) dy = frac {d}{dx}int_0^{infty} e^{-y(1+x)}y dy$
And one more time.
$I(x) = frac {d^2}{dx^2}int_0^{infty} e^{-y(1+x)} dy = frac {d^2}{dx^2}frac {1}{1+x} = frac {1}{2(1+x)^3}$
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$I(x) = int_0^{infty} e^{-y}e^{-xy}y^2 dy$
Integration under the integral
$I(x) = frac {d}{dx} int_0^{infty} e^{-y}y^2(int e^{-xy} dx) dy = frac {d}{dx}int_0^{infty} e^{-y(1+x)}y dy$
And one more time.
$I(x) = frac {d^2}{dx^2}int_0^{infty} e^{-y(1+x)} dy = frac {d^2}{dx^2}frac {1}{1+x} = frac {1}{2(1+x)^3}$
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
$endgroup$
$I(x) = int_0^{infty} e^{-y}e^{-xy}y^2 dy$
Integration under the integral
$I(x) = frac {d}{dx} int_0^{infty} e^{-y}y^2(int e^{-xy} dx) dy = frac {d}{dx}int_0^{infty} e^{-y(1+x)}y dy$
And one more time.
$I(x) = frac {d^2}{dx^2}int_0^{infty} e^{-y(1+x)} dy = frac {d^2}{dx^2}frac {1}{1+x} = frac {1}{2(1+x)^3}$
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
answered Jan 23 at 21:27
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
$begingroup$
Try separating the integral like this
$$int _0^infty e^{-y(x+1)}y^2dy=int _0^1 e^{-y(x+1)}y^2dy+int _1^infty e^{-y(x+1)}y^2dy$$
you can see that the first integral is a proper one hence the integral converges.
For the second one try taking the cases when $x+1 lt0 $, $x+1 =0$ and $x+1 gt0$
for $x+1 gt0$ from $$lim_{yto infty}frac{e^{-y(x+1)}y^2}{frac{1}{y2}}=0$$ this implies that $$int _1^infty e^{-y(x+1)}y^2dyleint _1^infty frac{1}{y^2}dy$$
the last integral converges, from the comparison test so will the integral $int _1^infty e^{-y(x+1)}y^2dy$
Can you try the other cases?
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
$endgroup$
add a comment |
$begingroup$
Try separating the integral like this
$$int _0^infty e^{-y(x+1)}y^2dy=int _0^1 e^{-y(x+1)}y^2dy+int _1^infty e^{-y(x+1)}y^2dy$$
you can see that the first integral is a proper one hence the integral converges.
For the second one try taking the cases when $x+1 lt0 $, $x+1 =0$ and $x+1 gt0$
for $x+1 gt0$ from $$lim_{yto infty}frac{e^{-y(x+1)}y^2}{frac{1}{y2}}=0$$ this implies that $$int _1^infty e^{-y(x+1)}y^2dyleint _1^infty frac{1}{y^2}dy$$
the last integral converges, from the comparison test so will the integral $int _1^infty e^{-y(x+1)}y^2dy$
Can you try the other cases?
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
$endgroup$
add a comment |
$begingroup$
Try separating the integral like this
$$int _0^infty e^{-y(x+1)}y^2dy=int _0^1 e^{-y(x+1)}y^2dy+int _1^infty e^{-y(x+1)}y^2dy$$
you can see that the first integral is a proper one hence the integral converges.
For the second one try taking the cases when $x+1 lt0 $, $x+1 =0$ and $x+1 gt0$
for $x+1 gt0$ from $$lim_{yto infty}frac{e^{-y(x+1)}y^2}{frac{1}{y2}}=0$$ this implies that $$int _1^infty e^{-y(x+1)}y^2dyleint _1^infty frac{1}{y^2}dy$$
the last integral converges, from the comparison test so will the integral $int _1^infty e^{-y(x+1)}y^2dy$
Can you try the other cases?
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
$endgroup$
Try separating the integral like this
$$int _0^infty e^{-y(x+1)}y^2dy=int _0^1 e^{-y(x+1)}y^2dy+int _1^infty e^{-y(x+1)}y^2dy$$
you can see that the first integral is a proper one hence the integral converges.
For the second one try taking the cases when $x+1 lt0 $, $x+1 =0$ and $x+1 gt0$
for $x+1 gt0$ from $$lim_{yto infty}frac{e^{-y(x+1)}y^2}{frac{1}{y2}}=0$$ this implies that $$int _1^infty e^{-y(x+1)}y^2dyleint _1^infty frac{1}{y^2}dy$$
the last integral converges, from the comparison test so will the integral $int _1^infty e^{-y(x+1)}y^2dy$
Can you try the other cases?
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
edited Jan 23 at 21:34
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
answered Jan 23 at 21:24
J.DaneJ.Dane
368114
368114
add a comment |
add a comment |
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$begingroup$
This integral looks very definite...
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– user
Jan 23 at 21:26
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@user I edited it to include limits to match the stated result. I'll fix the title.
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– J.G.
Jan 23 at 21:31