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Find $k$ value so the function is a pdf

(a) $kx^6(1-x)^4$, for $0 < x < 1, 0$ otherwise

beta distribution has this formula

$f(\alpha, \beta) = x^{\alpha-1}(1-x)^{\beta-1} = \frac{T(\alpha + \beta)}{T(\alpha)T(\beta)}$

Attempt

$\alpha = 7$

$\beta = 5$

Not sure how to apply this

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    $\begingroup$ $k$ is the normalizing factor. When you integrate the pdf in its domain, what should be its value? $\endgroup$ – karakfa Jun 13 '18 at 21:35
  • $\begingroup$ Integrating it would require me to expand $(1-x)^4$ or do like 6 integration by parts alternatively we've been told to do this using beta distribution which I do not understand. $\endgroup$ – Bas bas Jun 13 '18 at 22:00
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Hint:

$$ \int_0^1 x^{\alpha-1}(1-x)^{\beta-1} = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} = B(\alpha,\beta) $$

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  • $\begingroup$ What is $$\frac{T(7)T(T5)}{T(12)}$$ I do not know what T is $\endgroup$ – Bas bas Jun 13 '18 at 22:13
  • $\begingroup$ It's gamma function, not $T$. If you want to do shortcuts you need to know shortcuts. Otherwise the integral is not difficult to calculate for integer $\alpha, \beta$. $\endgroup$ – karakfa Jun 14 '18 at 0:13
  • $\begingroup$ google says its T(x) = (x-1)!. So this would equal 1/2310, therefore k = 2310 . dunno. $\endgroup$ – Bas bas Jun 14 '18 at 0:26

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