Relate exponential to normal distribution and find $E[X^2]$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Expectation of the maximum of gaussian random variablesHow to find probability density functions?Probability density function of the integral of a continuous stochastic processCalculating the variance and standard deviation of a Laplace distributionHow to find the pdf of difference of r.vSquare of Normal distributed variable?Coin Toss Question with Normal Distribution.Compound normal distribution with mean from truncated normalCombination of Normal random variable and BernoulliPolar form of normal random vector , angle and length are independent ,and angle is spherical distribution
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Relate exponential to normal distribution and find $E[X^2]$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Expectation of the maximum of gaussian random variablesHow to find probability density functions?Probability density function of the integral of a continuous stochastic processCalculating the variance and standard deviation of a Laplace distributionHow to find the pdf of difference of r.vSquare of Normal distributed variable?Coin Toss Question with Normal Distribution.Compound normal distribution with mean from truncated normalCombination of Normal random variable and BernoulliPolar form of normal random vector , angle and length are independent ,and angle is spherical distribution
$begingroup$
The following question is absolutely killing me:
Let $X$ be a continuous r.v. distributed according to the pdf $ke^-x^2-7x$. Find $E[X^2]$.
I can supposedly map this onto the Gaussian pdf and use the variance equation to get the answer (i.e., allegedly no integrals required), however, I tried a one-to-one mapping of this pdf onto the Gaussian and came up with an embarrassing mess of algebra. So, I'm not at all sure how the suggested solution is supposed to work. Anyone else understand how to solve this?
probability
asked Apr 1 at 13:33
RyanRyan
1277
$endgroup$
add a comment |
$begingroup$
The following question is absolutely killing me:
Let $X$ be a continuous r.v. distributed according to the pdf $ke^-x^2-7x$. Find $E[X^2]$.
I can supposedly map this onto the Gaussian pdf and use the variance equation to get the answer (i.e., allegedly no integrals required), however, I tried a one-to-one mapping of this pdf onto the Gaussian and came up with an embarrassing mess of algebra. So, I'm not at all sure how the suggested solution is supposed to work. Anyone else understand how to solve this?
probability
asked Apr 1 at 13:33
RyanRyan
1277
$endgroup$
$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
1
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
1
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51
add a comment |
$begingroup$
The following question is absolutely killing me:
Let $X$ be a continuous r.v. distributed according to the pdf $ke^-x^2-7x$. Find $E[X^2]$.
I can supposedly map this onto the Gaussian pdf and use the variance equation to get the answer (i.e., allegedly no integrals required), however, I tried a one-to-one mapping of this pdf onto the Gaussian and came up with an embarrassing mess of algebra. So, I'm not at all sure how the suggested solution is supposed to work. Anyone else understand how to solve this?
probability
asked Apr 1 at 13:33
RyanRyan
1277
$endgroup$
The following question is absolutely killing me:
Let $X$ be a continuous r.v. distributed according to the pdf $ke^-x^2-7x$. Find $E[X^2]$.
I can supposedly map this onto the Gaussian pdf and use the variance equation to get the answer (i.e., allegedly no integrals required), however, I tried a one-to-one mapping of this pdf onto the Gaussian and came up with an embarrassing mess of algebra. So, I'm not at all sure how the suggested solution is supposed to work. Anyone else understand how to solve this?
probability
probability
asked Apr 1 at 13:33
RyanRyan
1277
asked Apr 1 at 13:33
RyanRyan
1277
asked Apr 1 at 13:33
RyanRyan
1277
asked Apr 1 at 13:33
RyanRyan
1277
asked Apr 1 at 13:33
RyanRyan
1277
1277
$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
1
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
1
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51
add a comment |
$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
1
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
1
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51
$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
1
1
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
1
1
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's complete the square. We have
beginalign*
kexp(-x^2 - 7x) &= kexpbig(-x^2 - 7x - frac494 + frac494big) \
&= k' expbigg(-big(x+frac72big)^2bigg).
endalign*
where $k' = kexpbig(frac494big)$. We see we have a normal distribution with mean $mu = -frac72$ and variance $sigma = sqrt2$. Then, like you said in the problem statement, we can use the fact that
$$
sigma^2 = mathbbE(X^2) - mathbbE(X)^2
$$
to get
$$
2 = mathbbE(X^2) - frac494
$$
or
$$
mathbbE(X^2) = frac574.
$$
answered Apr 1 at 13:43
hunterhunter
15.9k32643
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's complete the square. We have
beginalign*
kexp(-x^2 - 7x) &= kexpbig(-x^2 - 7x - frac494 + frac494big) \
&= k' expbigg(-big(x+frac72big)^2bigg).
endalign*
where $k' = kexpbig(frac494big)$. We see we have a normal distribution with mean $mu = -frac72$ and variance $sigma = sqrt2$. Then, like you said in the problem statement, we can use the fact that
$$
sigma^2 = mathbbE(X^2) - mathbbE(X)^2
$$
to get
$$
2 = mathbbE(X^2) - frac494
$$
or
$$
mathbbE(X^2) = frac574.
$$
answered Apr 1 at 13:43
hunterhunter
15.9k32643
$endgroup$
add a comment |
$begingroup$
Let's complete the square. We have
beginalign*
kexp(-x^2 - 7x) &= kexpbig(-x^2 - 7x - frac494 + frac494big) \
&= k' expbigg(-big(x+frac72big)^2bigg).
endalign*
where $k' = kexpbig(frac494big)$. We see we have a normal distribution with mean $mu = -frac72$ and variance $sigma = sqrt2$. Then, like you said in the problem statement, we can use the fact that
$$
sigma^2 = mathbbE(X^2) - mathbbE(X)^2
$$
to get
$$
2 = mathbbE(X^2) - frac494
$$
or
$$
mathbbE(X^2) = frac574.
$$
answered Apr 1 at 13:43
hunterhunter
15.9k32643
$endgroup$
add a comment |
$begingroup$
Let's complete the square. We have
beginalign*
kexp(-x^2 - 7x) &= kexpbig(-x^2 - 7x - frac494 + frac494big) \
&= k' expbigg(-big(x+frac72big)^2bigg).
endalign*
where $k' = kexpbig(frac494big)$. We see we have a normal distribution with mean $mu = -frac72$ and variance $sigma = sqrt2$. Then, like you said in the problem statement, we can use the fact that
$$
sigma^2 = mathbbE(X^2) - mathbbE(X)^2
$$
to get
$$
2 = mathbbE(X^2) - frac494
$$
or
$$
mathbbE(X^2) = frac574.
$$
answered Apr 1 at 13:43
hunterhunter
15.9k32643
$endgroup$
Let's complete the square. We have
beginalign*
kexp(-x^2 - 7x) &= kexpbig(-x^2 - 7x - frac494 + frac494big) \
&= k' expbigg(-big(x+frac72big)^2bigg).
endalign*
where $k' = kexpbig(frac494big)$. We see we have a normal distribution with mean $mu = -frac72$ and variance $sigma = sqrt2$. Then, like you said in the problem statement, we can use the fact that
$$
sigma^2 = mathbbE(X^2) - mathbbE(X)^2
$$
to get
$$
2 = mathbbE(X^2) - frac494
$$
or
$$
mathbbE(X^2) = frac574.
$$
answered Apr 1 at 13:43
hunterhunter
15.9k32643
answered Apr 1 at 13:43
hunterhunter
15.9k32643
answered Apr 1 at 13:43
hunterhunter
15.9k32643
answered Apr 1 at 13:43
hunterhunter
15.9k32643
15.9k32643
add a comment |
add a comment |
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$begingroup$
Firstly you have to find the value of k. Have you done that? And what is the domain of $x$?
$endgroup$
– callculus
Apr 1 at 13:37
1
$begingroup$
Hint: use complete the square
$endgroup$
– George Dewhirst
Apr 1 at 13:43
1
$begingroup$
I don't think it is necessary to find $k$.
$endgroup$
– hunter
Apr 1 at 13:44
$begingroup$
@hunter I think you´re right.
$endgroup$
– callculus
Apr 1 at 13:51