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Question: Evaluate: $$ \int_{0}^{1}x^m(1-x)^n dx$$ where $m, n$ are positive integers.

The hint is to interpret the probability as an integral.

What I've tried: For a fixed $x \in [0,1]$, $x^m(1-x)^n$ is the probability that you get $m$ heads followed by $n$ tails when you toss a coin (with heads probability $x$) $m+n$ times. Another way to write this is: $$P(\text{$m$ heads followed by $n$ heads } | \text{ Get $m$ heads within the $m+n$ tosses}) \times P(\text{Get $m$ heads within the $m+n$ tosses}) $$

Notice that if you know that there are $m$ heads, each of the ${m +n \choose n}$ outcomes are equally likely so we can reduce the above product of probabilities to: $\frac{1}{{m +n \choose n}} \cdot P(\text{Get $m$ heads within the $m+n$ tosses})$.

Now suppose $x$ is not fixed but rather chosen uniformly at random from $[0,1]$, then the probability that we get $m$ heads followed by $n$ tails is exactly the integral in the question above. Due to the discussion in the paragraph above, we can write this as:

$$\frac{1}{{m +n \choose n}} \int_{0}^{1} P(\text{Get $m$ heads within the $m+n$ tosses } | \text{ Heads probability is $x$}) dx$$ I do not know how to evaluate the above integral, however some numerical trials on WolframAlpha indicate that it should be $\frac{1}{m+n+1}$ giving a final answer of $\frac{1}{(m+n+1){m +n \choose n}}$.

I have no clue if this is on the right track and would appreciate any help.

basketball9
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  • The final answer looks correct. You can also solve this using the binomial theorem and some more binomial sum identities. – ljeabmreosn Mar 20 '20 at 19:31

2 Answers2

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This is essentially the same question as Why are all subset sizes equiprobable if elements are independently included with probability uniform over $[0,1]$?, but coming at it from the other side.

You were indeed on the right track with a probability uniformly randomly drawn from $[0,1]$. The clue to getting the factor $m+n+1$ without explicit integration is to consider the coin tosses as drawing from that same uniform distribution: Initially a probability $p$ is drawn uniformly from $[0,1]$, and then for each coin toss a number $r$ is drawn uniformly from $[0,1]$ and compared to $p$. The result is heads exactly if $r\lt p$. For $k$ tosses, that’s $k+1$ draws from the uniform distribution, and by symmetry $p$ is equally likely to have any of the $k+1$ ranks among these $k+1$ numbers. Thus each number of heads from $0$ to $k$ is equally likely. The probability for $m$ heads and $k-m$ tails is, as you found,

$$ \int_0^1\binom kmx^m(1-x)^{k-m}\mathrm dx\;, $$

and by the above, this must be the same for all values of $m$, and thus $\frac1{k+1}$. So indeed the value of your integral is

$$ \int_0^1x^m(1-x)^n\mathrm dx=\frac1{\binom{m+n}n(m+n+1)}\;. $$

joriki
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Hint: If $F(x)$ is a cumulative distribution function, then $lim_{x\to +\infty}F(x) =1$

Does this specific integral remind you of any specific continuous probability density function?

$$f(x; a; b) = \frac{t^{a-1}(1-t)^{b-1}}{B(a, b)}$$

is the pdf of the Beta distribution

Since $F(1) = \int_0^1f(x;a;b)=1$ you can multiply your given integral with a certain constant and then easily find the answer

Eleftheria Chatziargyriou
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