Finding Density Function of a Function of other Random Variables
$begingroup$
Suppose that two electronic components in the guidance system for a missile operate indepen-
dently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.
My solution:
Since $lambda = 1$, $f(y_1) = e^{-y_1}$ and $f(y_2) = e^{-y_2}$ for $y_1 > 0, y_2 > 0$. Since they operate independently, we have
$$
f(y_1,y_2) = f(y_1)f(y_2) = e^{-y_1} e^{-y_2}
$$
The random variable of interest is $U = frac{Y_1 + Y_2}{2}$. As such, we seek the distribution function $F_U(u) = f_U(U < u) = f_U(frac{Y_1 + Y_2}{2} < u)$
$$
= iint_{frac{Y_1 + Y_2}{2}<u}f(y_1,y_2)dy_1dy_2
$$
$$
= int_0^{2u} int_0^{2u-y_1}e^{-y_1} e^{-y_2} dy_2dy_1 = -e^{-2u} -2ue^{-2u} +1
$$
Therefore, we have that
$$
F_U(u) = begin{cases}
0 & u < 0\
-e^{-2u} -2ue^{-2u} +1 & u geq 0\
end{cases}
$$
therefore, for $u geq 0$,
$$
f_U(u) = frac{dF_U(u)}{du} = 2e^{-2u} + 4ue^{-2u}
$$
The answer, however, is $4ue^{-2u}$. I don't understand where the extra $2e^{-2u}$ came from. Where did I go wrong?
probability-distributions
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
$endgroup$
add a comment |
$begingroup$
Suppose that two electronic components in the guidance system for a missile operate indepen-
dently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.
My solution:
Since $lambda = 1$, $f(y_1) = e^{-y_1}$ and $f(y_2) = e^{-y_2}$ for $y_1 > 0, y_2 > 0$. Since they operate independently, we have
$$
f(y_1,y_2) = f(y_1)f(y_2) = e^{-y_1} e^{-y_2}
$$
The random variable of interest is $U = frac{Y_1 + Y_2}{2}$. As such, we seek the distribution function $F_U(u) = f_U(U < u) = f_U(frac{Y_1 + Y_2}{2} < u)$
$$
= iint_{frac{Y_1 + Y_2}{2}<u}f(y_1,y_2)dy_1dy_2
$$
$$
= int_0^{2u} int_0^{2u-y_1}e^{-y_1} e^{-y_2} dy_2dy_1 = -e^{-2u} -2ue^{-2u} +1
$$
Therefore, we have that
$$
F_U(u) = begin{cases}
0 & u < 0\
-e^{-2u} -2ue^{-2u} +1 & u geq 0\
end{cases}
$$
therefore, for $u geq 0$,
$$
f_U(u) = frac{dF_U(u)}{du} = 2e^{-2u} + 4ue^{-2u}
$$
The answer, however, is $4ue^{-2u}$. I don't understand where the extra $2e^{-2u}$ came from. Where did I go wrong?
probability-distributions
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
$endgroup$
add a comment |
$begingroup$
Suppose that two electronic components in the guidance system for a missile operate indepen-
dently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.
My solution:
Since $lambda = 1$, $f(y_1) = e^{-y_1}$ and $f(y_2) = e^{-y_2}$ for $y_1 > 0, y_2 > 0$. Since they operate independently, we have
$$
f(y_1,y_2) = f(y_1)f(y_2) = e^{-y_1} e^{-y_2}
$$
The random variable of interest is $U = frac{Y_1 + Y_2}{2}$. As such, we seek the distribution function $F_U(u) = f_U(U < u) = f_U(frac{Y_1 + Y_2}{2} < u)$
$$
= iint_{frac{Y_1 + Y_2}{2}<u}f(y_1,y_2)dy_1dy_2
$$
$$
= int_0^{2u} int_0^{2u-y_1}e^{-y_1} e^{-y_2} dy_2dy_1 = -e^{-2u} -2ue^{-2u} +1
$$
Therefore, we have that
$$
F_U(u) = begin{cases}
0 & u < 0\
-e^{-2u} -2ue^{-2u} +1 & u geq 0\
end{cases}
$$
therefore, for $u geq 0$,
$$
f_U(u) = frac{dF_U(u)}{du} = 2e^{-2u} + 4ue^{-2u}
$$
The answer, however, is $4ue^{-2u}$. I don't understand where the extra $2e^{-2u}$ came from. Where did I go wrong?
probability-distributions
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
$endgroup$
Suppose that two electronic components in the guidance system for a missile operate indepen-
dently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.
My solution:
Since $lambda = 1$, $f(y_1) = e^{-y_1}$ and $f(y_2) = e^{-y_2}$ for $y_1 > 0, y_2 > 0$. Since they operate independently, we have
$$
f(y_1,y_2) = f(y_1)f(y_2) = e^{-y_1} e^{-y_2}
$$
The random variable of interest is $U = frac{Y_1 + Y_2}{2}$. As such, we seek the distribution function $F_U(u) = f_U(U < u) = f_U(frac{Y_1 + Y_2}{2} < u)$
$$
= iint_{frac{Y_1 + Y_2}{2}<u}f(y_1,y_2)dy_1dy_2
$$
$$
= int_0^{2u} int_0^{2u-y_1}e^{-y_1} e^{-y_2} dy_2dy_1 = -e^{-2u} -2ue^{-2u} +1
$$
Therefore, we have that
$$
F_U(u) = begin{cases}
0 & u < 0\
-e^{-2u} -2ue^{-2u} +1 & u geq 0\
end{cases}
$$
therefore, for $u geq 0$,
$$
f_U(u) = frac{dF_U(u)}{du} = 2e^{-2u} + 4ue^{-2u}
$$
The answer, however, is $4ue^{-2u}$. I don't understand where the extra $2e^{-2u}$ came from. Where did I go wrong?
probability-distributions
probability-distributions
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
asked Dec 30 '18 at 4:32
Bryden CBryden C
30418
30418
add a comment |
add a comment |
1 Answer
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I think you computed the derivative incorrectly.
$$frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
$endgroup$
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think you computed the derivative incorrectly.
$$frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
$endgroup$
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
add a comment |
$begingroup$
I think you computed the derivative incorrectly.
$$frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
$endgroup$
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
add a comment |
$begingroup$
I think you computed the derivative incorrectly.
$$frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
$endgroup$
I think you computed the derivative incorrectly.
$$frac{d}{du} (-2ue^{-2u}) = 4ue^{-2u} - 2 e^{-2u}.$$
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
answered Dec 30 '18 at 4:37
angryavianangryavian
39.7k23280
39.7k23280
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
add a comment |
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
1
1
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
$begingroup$
Wow, I feel really silly right about now. That would explain a lot. At least I didn't make an error setting up the distribution function.
$endgroup$
– Bryden C
Dec 30 '18 at 4:39
add a comment |
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