$ f(x) = \begin{cases} k\sqrt{x}, 0 I know that $E[X] = \frac{2k}{5}$ and $Var[X] = \frac {2k}{7}$ Then what can I do to find $k$ with these few information?
Find a constant in a density function
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probability-distributions
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0oops...I approve the edit by Rick wrongly. – 2012-10-23
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0However, there's a typo mistake, it should be $0
instead of $x>0$ – 2012-10-23
1 Answers
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Updated to match the corrected version of the question:
You must have $$\int_{-\infty}^\infty f(x)~dx=1$$ in order for $f$ to be a probability density function. In this case
$$\int_{-\infty}^\infty f(x)~dx=\int_0^1 k\sqrt x~dx=k\int_0^1 x^{1/2}~dx\;,$$
so you need only solve the equation
$$k\int_0^1 x^{1/2}~dx=1$$
for $k$.
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0Sorry sir, there's a typo mistake, it should be $0
instead of $x>0$ in the first line. – 2012-10-23 -
0If it's $0
, is $k=\frac32 $? – 2012-10-23 -
0@Justin: Yes, in that case $k=\frac32$; I’ll update the answer. – 2012-10-23
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0I got it. Thanks bro. – 2012-10-23
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0@Justin: You’re welcome. – 2012-10-23
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0@Justin: maybe you should verify $Var[X] = \frac{2k}{7}$ to make sure the problem makes sense. – 2012-10-23