Mixing
Naive question about continuous random variable: solution verification
$begingroup$
The probability density function of the random variable $X$ is given by
$$f(x) =
begin{cases}
a + bx^{2} & 0 < x < 1 \
0 & text{otherwise}
end{cases}$$
If $textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
begin{align*}
int_{0}^{1}(a + bx^{2})mathrm{d}x = a + frac{b}{3} = 1
end{align*}
On the other hand, according to the definition of expected value, we get
begin{align*}
int_{0}^{1}(ax+ bx^{3})mathrm{d}x = frac{a}{2} + frac{b}{4} = frac{3}{5}
end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.
proof-verification density-function
asked Jan 31 at 22:40
APC89APC89
2,371720
$endgroup$
add a comment |
$begingroup$
The probability density function of the random variable $X$ is given by
$$f(x) =
begin{cases}
a + bx^{2} & 0 < x < 1 \
0 & text{otherwise}
end{cases}$$
If $textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
begin{align*}
int_{0}^{1}(a + bx^{2})mathrm{d}x = a + frac{b}{3} = 1
end{align*}
On the other hand, according to the definition of expected value, we get
begin{align*}
int_{0}^{1}(ax+ bx^{3})mathrm{d}x = frac{a}{2} + frac{b}{4} = frac{3}{5}
end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.
proof-verification density-function
asked Jan 31 at 22:40
APC89APC89
2,371720
$endgroup$
$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51
add a comment |
$begingroup$
The probability density function of the random variable $X$ is given by
$$f(x) =
begin{cases}
a + bx^{2} & 0 < x < 1 \
0 & text{otherwise}
end{cases}$$
If $textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
begin{align*}
int_{0}^{1}(a + bx^{2})mathrm{d}x = a + frac{b}{3} = 1
end{align*}
On the other hand, according to the definition of expected value, we get
begin{align*}
int_{0}^{1}(ax+ bx^{3})mathrm{d}x = frac{a}{2} + frac{b}{4} = frac{3}{5}
end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.
proof-verification density-function
asked Jan 31 at 22:40
APC89APC89
2,371720
$endgroup$
The probability density function of the random variable $X$ is given by
$$f(x) =
begin{cases}
a + bx^{2} & 0 < x < 1 \
0 & text{otherwise}
end{cases}$$
If $textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
begin{align*}
int_{0}^{1}(a + bx^{2})mathrm{d}x = a + frac{b}{3} = 1
end{align*}
On the other hand, according to the definition of expected value, we get
begin{align*}
int_{0}^{1}(ax+ bx^{3})mathrm{d}x = frac{a}{2} + frac{b}{4} = frac{3}{5}
end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.
proof-verification density-function
proof-verification density-function
asked Jan 31 at 22:40
APC89APC89
2,371720
asked Jan 31 at 22:40
APC89APC89
2,371720
asked Jan 31 at 22:40
APC89APC89
2,371720
asked Jan 31 at 22:40
APC89APC89
2,371720
asked Jan 31 at 22:40
APC89APC89
2,371720
2,371720
$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51
add a comment |
$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51
$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Everything's correct; solution is short, but totally understandable!
answered Jan 31 at 22:45
IngixIngix
5,162159
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Everything's correct; solution is short, but totally understandable!
answered Jan 31 at 22:45
IngixIngix
5,162159
$endgroup$
add a comment |
$begingroup$
Everything's correct; solution is short, but totally understandable!
answered Jan 31 at 22:45
IngixIngix
5,162159
$endgroup$
add a comment |
$begingroup$
Everything's correct; solution is short, but totally understandable!
answered Jan 31 at 22:45
IngixIngix
5,162159
$endgroup$
Everything's correct; solution is short, but totally understandable!
answered Jan 31 at 22:45
IngixIngix
5,162159
answered Jan 31 at 22:45
IngixIngix
5,162159
answered Jan 31 at 22:45
IngixIngix
5,162159
answered Jan 31 at 22:45
IngixIngix
5,162159
5,162159
add a comment |
add a comment |
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$begingroup$
It's fine. Any doubts in particular?
$endgroup$
– Alejandro Nasif Salum
Jan 31 at 22:44
$begingroup$
Nothing in particular. I was just asking for double-checking. Thanks by the way.
$endgroup$
– APC89
Jan 31 at 22:51