Mixing
Gamma distribution getting predictions
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
frac{81}{8}x^4e^{-3x}& x > 0 \
0 & otherwise \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
attempt
I know this is the gamma distribution
(a)
This is $X$ ~ $Gamma(a=5, lambda=3)$ so $E(X) = frac{5}{3}$. By definition of Gamma $E(X) = frac{a}{lambda}$
(b) $f(x) = frac{81}{8}x^4e^{-3x}dx$
$f'(x) = frac{81}{8}x^3e^{-3x}(4-3x) = 0$, so $x = 0, 4/3$
The maximum is given by $x = 4/3$ in the original equation.
$f(4/3) approx 0.586$
(c)
$MSE(mode) = E(X - frac{4}{3})^2 = E(X^2 - frac{8}{3}X + 16/9)$
$E(X^2) - 8/3E(X) + 16/9$
Need to find $E(X^2)$. We know $V(X) = E(X^2) - E(X)^2 leftrightarrow E(X^2) = V(X) + E(X)^2$ and $V(X) = frac{a}{lambda^2} = frac{5}{9}$ for gamma and $E(X) = frac{5}{3}$
so $E(X^2) = 5/9 + (5/3)^2 = 10/3$
therefore $MSE(Mode) = E(X^2) - 8/3E(X) + 16/9 = 10/3 - 8/3 cdot 5/3 + 16/9 = 2/3$
Is this correct?
probability probability-distributions
asked Jan 16 at 20:51
shahshah
404
$endgroup$
add a comment |
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
frac{81}{8}x^4e^{-3x}& x > 0 \
0 & otherwise \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
attempt
I know this is the gamma distribution
(a)
This is $X$ ~ $Gamma(a=5, lambda=3)$ so $E(X) = frac{5}{3}$. By definition of Gamma $E(X) = frac{a}{lambda}$
(b) $f(x) = frac{81}{8}x^4e^{-3x}dx$
$f'(x) = frac{81}{8}x^3e^{-3x}(4-3x) = 0$, so $x = 0, 4/3$
The maximum is given by $x = 4/3$ in the original equation.
$f(4/3) approx 0.586$
(c)
$MSE(mode) = E(X - frac{4}{3})^2 = E(X^2 - frac{8}{3}X + 16/9)$
$E(X^2) - 8/3E(X) + 16/9$
Need to find $E(X^2)$. We know $V(X) = E(X^2) - E(X)^2 leftrightarrow E(X^2) = V(X) + E(X)^2$ and $V(X) = frac{a}{lambda^2} = frac{5}{9}$ for gamma and $E(X) = frac{5}{3}$
so $E(X^2) = 5/9 + (5/3)^2 = 10/3$
therefore $MSE(Mode) = E(X^2) - 8/3E(X) + 16/9 = 10/3 - 8/3 cdot 5/3 + 16/9 = 2/3$
Is this correct?
probability probability-distributions
asked Jan 16 at 20:51
shahshah
404
$endgroup$
add a comment |
$begingroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
frac{81}{8}x^4e^{-3x}& x > 0 \
0 & otherwise \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
attempt
I know this is the gamma distribution
(a)
This is $X$ ~ $Gamma(a=5, lambda=3)$ so $E(X) = frac{5}{3}$. By definition of Gamma $E(X) = frac{a}{lambda}$
(b) $f(x) = frac{81}{8}x^4e^{-3x}dx$
$f'(x) = frac{81}{8}x^3e^{-3x}(4-3x) = 0$, so $x = 0, 4/3$
The maximum is given by $x = 4/3$ in the original equation.
$f(4/3) approx 0.586$
(c)
$MSE(mode) = E(X - frac{4}{3})^2 = E(X^2 - frac{8}{3}X + 16/9)$
$E(X^2) - 8/3E(X) + 16/9$
Need to find $E(X^2)$. We know $V(X) = E(X^2) - E(X)^2 leftrightarrow E(X^2) = V(X) + E(X)^2$ and $V(X) = frac{a}{lambda^2} = frac{5}{9}$ for gamma and $E(X) = frac{5}{3}$
so $E(X^2) = 5/9 + (5/3)^2 = 10/3$
therefore $MSE(Mode) = E(X^2) - 8/3E(X) + 16/9 = 10/3 - 8/3 cdot 5/3 + 16/9 = 2/3$
Is this correct?
probability probability-distributions
asked Jan 16 at 20:51
shahshah
404
$endgroup$
Random variable $X$ has a continuous distribution with the probability density function below
$$f_X(x) =
begin{cases}
frac{81}{8}x^4e^{-3x}& x > 0 \
0 & otherwise \
end{cases}
$$
(a) Determine the prediction of for a future $X$, if we have decided to predict using the mean of the distribution
(b) Determine the prediction of a future $X$, if we have decided to predict using the mode of the distribution
(c) Calculate the mean-squared error of the predictor in part(b) above.
attempt
I know this is the gamma distribution
(a)
This is $X$ ~ $Gamma(a=5, lambda=3)$ so $E(X) = frac{5}{3}$. By definition of Gamma $E(X) = frac{a}{lambda}$
(b) $f(x) = frac{81}{8}x^4e^{-3x}dx$
$f'(x) = frac{81}{8}x^3e^{-3x}(4-3x) = 0$, so $x = 0, 4/3$
The maximum is given by $x = 4/3$ in the original equation.
$f(4/3) approx 0.586$
(c)
$MSE(mode) = E(X - frac{4}{3})^2 = E(X^2 - frac{8}{3}X + 16/9)$
$E(X^2) - 8/3E(X) + 16/9$
Need to find $E(X^2)$. We know $V(X) = E(X^2) - E(X)^2 leftrightarrow E(X^2) = V(X) + E(X)^2$ and $V(X) = frac{a}{lambda^2} = frac{5}{9}$ for gamma and $E(X) = frac{5}{3}$
so $E(X^2) = 5/9 + (5/3)^2 = 10/3$
therefore $MSE(Mode) = E(X^2) - 8/3E(X) + 16/9 = 10/3 - 8/3 cdot 5/3 + 16/9 = 2/3$
Is this correct?
probability probability-distributions
probability probability-distributions
asked Jan 16 at 20:51
shahshah
404
asked Jan 16 at 20:51
shahshah
404
asked Jan 16 at 20:51
shahshah
404
asked Jan 16 at 20:51
shahshah
404
asked Jan 16 at 20:51
shahshah
404
404
add a comment |
add a comment |
1 Answer
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$begingroup$
(a) You are correct to recognize that $X sim text{Gamma}(5, 3)$, and your expectation is also correct.
(b) Your mode is correct. A quicker way to do it, however, is to recognize that the mode of random variable following a Gamma distribution with parameters $alpha$ and $lambda$ is given by $(alpha - 1)/lambda$ for $alpha geq 1$. So, in this case, the mode would be at $(5 - 1)/3 = 4/3$, which agrees with what you have.
(c) Yes, it is correct.
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
add a comment |
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1 Answer
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$begingroup$
(a) You are correct to recognize that $X sim text{Gamma}(5, 3)$, and your expectation is also correct.
(b) Your mode is correct. A quicker way to do it, however, is to recognize that the mode of random variable following a Gamma distribution with parameters $alpha$ and $lambda$ is given by $(alpha - 1)/lambda$ for $alpha geq 1$. So, in this case, the mode would be at $(5 - 1)/3 = 4/3$, which agrees with what you have.
(c) Yes, it is correct.
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
add a comment |
$begingroup$
(a) You are correct to recognize that $X sim text{Gamma}(5, 3)$, and your expectation is also correct.
(b) Your mode is correct. A quicker way to do it, however, is to recognize that the mode of random variable following a Gamma distribution with parameters $alpha$ and $lambda$ is given by $(alpha - 1)/lambda$ for $alpha geq 1$. So, in this case, the mode would be at $(5 - 1)/3 = 4/3$, which agrees with what you have.
(c) Yes, it is correct.
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
add a comment |
$begingroup$
(a) You are correct to recognize that $X sim text{Gamma}(5, 3)$, and your expectation is also correct.
(b) Your mode is correct. A quicker way to do it, however, is to recognize that the mode of random variable following a Gamma distribution with parameters $alpha$ and $lambda$ is given by $(alpha - 1)/lambda$ for $alpha geq 1$. So, in this case, the mode would be at $(5 - 1)/3 = 4/3$, which agrees with what you have.
(c) Yes, it is correct.
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
$endgroup$
(a) You are correct to recognize that $X sim text{Gamma}(5, 3)$, and your expectation is also correct.
(b) Your mode is correct. A quicker way to do it, however, is to recognize that the mode of random variable following a Gamma distribution with parameters $alpha$ and $lambda$ is given by $(alpha - 1)/lambda$ for $alpha geq 1$. So, in this case, the mode would be at $(5 - 1)/3 = 4/3$, which agrees with what you have.
(c) Yes, it is correct.
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
answered Jan 16 at 20:58
Ekesh KumarEkesh Kumar
1,00428
1,00428
add a comment |
add a comment |
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