0

$\sum_{x=0}^{500}\frac{x-2}{x!(500-x)!}(0.4)^x(0.6)^{500-x}$

How do I tackle this question?

fossdeep
  • 239

1 Answers1

1

Algebraic Method

Let

$$A=\sum_{x=0}^{500}\frac{x}{x!(500-x)!}(0.4)^x(0.6)^{500-x};\qquad B=\sum_{x=0}^{500}\frac{1}{x!(500-x)!}(0.4)^x(0.6)^{500-x}$$

So the sum $S=A-2B$.

At this stage, it is evident that we have two series that each strongly resemble a binomial series for positive integer exponents, for which the standard form we will aim for is:

$$\boxed{(a+b)^n=\sum_{k=0}^{n}{\frac{n!}{k!(n-k)!}a^kb^{n-k}}}$$

So through some algebraic manipulation

$$\begin{align} A&=0+\sum_{x=1}^{500}\frac{x}{x!(500-x)!}(0.4)^x(0.6)^{500-x} \\[2ex] &=\sum_{x=1}^{500}\frac{1}{(x-1)!\left(499-(x-1)\right)!}(0.4)(0.4)^{x-1}(0.6)^{499-(x-1)} \\[2ex] &=\sum_{q=0}^{499}\frac{1}{q!(499-q)!}(0.4)(0.4)^q(0.6)^{499-q}&(q=x-1) \\[2ex] &=\frac{0.4}{499!}\sum_{q=0}^{499}\frac{499!}{q!(499-q)!}(0.4)^q(0.6)^{499-q} \\[2ex] &=\frac{0.4}{499!}(0.4+0.6)^{499} \\[2ex] &=\frac{0.4}{499!} \end{align}$$

Similarly $$\begin{align} B&=\sum_{x=0}^{500}\frac{1}{x!(500-x)!}(0.4)^x(0.6)^{500-x} \\[2ex] &=\frac{1}{500!}\sum_{x=0}^{500}\frac{500!}{x!(500-x)!}(0.4)^x(0.6)^{500-x} \\[2ex] &=\frac{1}{500!}(0.4+0.6)^{500} \\[2ex] &=\frac{1}{500!} \end{align}$$

Therefore

$$S=\frac{0.4}{499!}-\frac{2}{500!}=\frac{200-2}{500!}=\frac{198}{500!}$$

Statistical Method

$$S=\frac{1}{500!}\sum_{x=0}^{500}(x-2)\frac{500!}{x!(500-x)!}(0.4)^x(0.6)^{500-x}=\frac{1}{500!}\sum_{x=0}^{500}(x-2)\binom{500}{x}(0.4)^x(0.6)^{500-x}$$

which we recognise the expected value of $(X-2)$ for a variable $X$ which has a binomial distribution $X \sim \mathrm{Binom}(n,p)$ with $n=500,p=0.4$. So

$$S=\frac{1}{500!}\mathsf{E}[X-2]=\frac{1}{500!}(\mathsf{E}[X]-2)=\frac{np-2}{500!}=\frac{(500)(0.4)-2}{500!}=\frac{198}{500!}$$

where we used the fact that the expected value of a binomially distributed variable is the mean $np$.

Marconius
  • 5,635
  • I need some time to take this all in. First, why did you replace the x's on the fifth line with x-1's? – fossdeep Oct 23 '15 at 02:26
  • @fossdeep - The $x$ in the numerator is an issue, so cancel it, i.e. $\frac{x!}{x}=(x-1)!$. Once this has been done, the bases of the factorial terms in the denominator add to 499 not 500 (and this is the major determinant of what we will get in binomial series form), so we will need to reindex the summation by setting $q=x-1$ and to do this we need to put quantities in terms of $x-1$. – Marconius Oct 23 '15 at 02:52
  • Ah I must say, the statistical method is much easier to understand. Thanks a lot. – fossdeep Oct 23 '15 at 20:25
  • You're welcome! I assure you that the statistical method was much easier to apply too :-) – Marconius Oct 24 '15 at 01:23
  • Do you mind taking a stab at this question? I've tried using the same technique (which has helped me solve the next three), but I'm unable to do it for this last question. And I don't really understand the solutions others have posted. I feel like they don't understand my question? http://math.stackexchange.com/questions/1494656/using-distribution-properties-to-find-the-exact-value-of-the-following-sum – fossdeep Oct 24 '15 at 04:54
  • @fossdeep - I don't believe I can add anything to AndréNicolas' good answer to that question other than some minor explanation. You have $\frac{x^2}{2^x}=x^2(\frac{1}{2})^x=x^2(1-\frac{1}{2})^{x-1}(\frac{1}{2})$ which now fits the form of the geometric distribution (with $p=\frac{1}{2}$) for $\Pr(X=k)$ (see https://en.wikipedia.org/wiki/Geometric_distribution). The other bits are the formulae for $E(X)$ and $Var(X)$ for the geometric distribution, and the standard relationship (for any distribution) between variance and the second-moment and first-moment: $\text{Var}(X)=E(X^2)-(E(X))^2$. – Marconius Oct 24 '15 at 21:15
  • Hmm alright thank you. – fossdeep Oct 24 '15 at 22:10
  • Actually your comment really helped. The equation you gave in the second line was exactly what I was looking for. More statistical related than the other responses I guess. Thanks again. – fossdeep Oct 25 '15 at 19:39