PastPaper.workedSolution
Part A
(a) Differentiating using the product rule:
\( f_n'(x) = n x^{n-1} e^{-x} - x^n e^{-x} = x^{n-1} e^{-x} (n - x) \).
Since \( x > 0 \) and \( e^{-x} > 0 \), \( f_n'(x) = 0 \implies x = n \).
For \( 0 < x < n \), \( f_n'(x) > 0 \), and for \( x > n \), \( f_n'(x) < 0 \). Thus, \( x = n \) is a unique local maximum.
(b) The \( y \)-coordinate is \( f_n(n) = n^n e^{-n} = \left(\frac{n}{e}\right)^n \).
Since \( \frac{n}{e} > 1 \) for all \( n \geq 3 \), the base of the exponent grows beyond 1, so \( \lim_{n \to \infty} \left(\frac{n}{e}\right)^n = \infty \).
(c) Finding the second derivative:
\( f_n''(x) = (n-1)x^{n-2}e^{-x}(n-x) - x^{n-1}e^{-x} \)
\( f_n''(x) = x^{n-2}e^{-x} [ (n-1)(n-x) - x ] = x^{n-2}e^{-x} [ x^2 - 2nx + n(n-1) ] \).
Setting \( f_n''(x) = 0 \) for \( x > 0 \) gives the quadratic equation:
\( x^2 - 2nx + n(n-1) = 0 \).
Using the quadratic formula:
\( x = \frac{2n \pm \sqrt{4n^2 - 4n(n-1)}}{2} = n \pm \sqrt{n} \).
Thus, the points of inflection have coordinates \( (n \pm \sqrt{n}, f_n(n \pm \sqrt{n})) \).
Part B
(d) \( I_0 = \int_0^\infty e^{-x} \, dx = [-e^{-x}]_0^\infty = 0 - (-1) = 1 \).
(e) Using integration by parts with \( u = x^n \implies du = n x^{n-1} dx \) and \( dv = e^{-x} dx \implies v = -e^{-x} \):
\( I_n = [-x^n e^{-x}]_0^\infty + n \int_0^\infty x^{n-1} e^{-x} \, dx \).
Since \( \lim_{x \to \infty} x^n e^{-x} = 0 \) and \( 0^n e^0 = 0 \), the boundary term vanishes.
Thus, \( I_n = n I_{n-1} \).
(f) Since \( I_0 = 1 \) and \( I_n = n I_{n-1} \), it follows that \( I_n = n! \).
(g) For \( n = 0 \):
LHS: \( J_0(t) = \int_0^t e^{-x} \, dx = 1 - e^{-t} \).
RHS: \( 0! \left( 1 - e^{-t} \frac{t^0}{0!} \right) = 1 - e^{-t} \). Thus, true for \( n = 0 \).
Assume the formula holds for \( n = k \): \( J_k(t) = k! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \right) \).
For \( n = k + 1 \):
\( J_{k+1}(t) = \int_0^t x^{k+1} e^{-x} \, dx \). Using integration by parts with \( u = x^{k+1} \) and \( dv = e^{-x} dx \):
\( J_{k+1}(t) = [-x^{k+1} e^{-x}]_0^t + (k+1) \int_0^t x^k e^{-x} \, dx = -t^{k+1} e^{-t} + (k+1) J_k(t) \).
Substituting the induction hypothesis:
\( J_{k+1}(t) = -t^{k+1} e^{-t} + (k+1) k! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \right) \)
\( = -t^{k+1} e^{-t} + (k+1)! - (k+1)! e^{-t} \sum_{r=0}^k \frac{t^r}{r!} \)
\( = (k+1)! \left( 1 - e^{-t} \sum_{r=0}^k \frac{t^r}{r!} - e^{-t} \frac{t^{k+1}}{(k+1)!} \right) \)
\( = (k+1)! \left( 1 - e^{-t} \sum_{r=0}^{k+1} \frac{t^r}{r!} \right) \).
Thus, by the principle of mathematical induction, the statement is true for all \( n \in \mathbb{N} \).
Part C
(h) \( \text{P}(X_t \leq n) = \sum_{k=0}^n \frac{e^{-t} t^k}{k!} \).
(i) From Part B, \( J_n(t) = n! - n! e^{-t} \sum_{k=0}^n \frac{t^k}{k!} \).
Since \( I_n = n! \), we can rewrite this as:
\( J_n(t) = I_n - n! \text{P}(X_t \leq n) \).
Thus, \( n! \text{P}(X_t \leq n) = I_n - J_n(t) = \int_0^\infty x^n e^{-x} \, dx - \int_0^t x^n e^{-x} \, dx = \int_t^\infty x^n e^{-x} \, dx \).
Dividing by \( n! \) gives: \( \text{P}(X_t \leq n) = \frac{1}{n!} \int_t^\infty x^n e^{-x} \, dx \).
(j) For large \( n \), a Poisson distribution with parameter \( t = n \) has mean \( \mu = n \) and variance \( \sigma^2 = n \).
By the Central Limit Theorem, \( X_n \) can be approximated by a normal distribution \( Y \sim \text{N}(n, n) \).
Therefore, \( \text{P}(X_n \leq n) \approx \text{P}(Y \leq n) = \text{P}\left(Z \leq \frac{n-n}{\sqrt{n}}\right) = \text{P}(Z \leq 0) = 0.5 \).
Thus, \( \lim_{n \to \infty} \text{P}(X_n \leq n) = 0.5 \).
(k) Since \( \frac{1}{n!} \int_n^\infty x^n e^{-x} \, dx = \text{P}(X_n \leq n) \), we have:
\( \lim_{n \to \infty} \frac{1}{n!} \int_n^\infty x^n e^{-x} \, dx = \lim_{n \to \infty} \text{P}(X_n \leq n) = 0.5 \).
PastPaper.markingScheme
Part A
(a) [2 marks]
M1: Attempt to differentiate \( f_n(x) \) using the product rule.
A1: Correct derivative set to 0 and obtaining \( x = n \) with valid reasoning for local maximum.
(b) [2 marks]
A1: Correct maximum coordinate \( \left(\frac{n}{e}\right)^n \).
A1: Correct justification of limit tending to \( \infty \).
(c) [2 marks]
M1: Setting second derivative to 0.
A1: Correct coordinates \( n \pm \sqrt{n} \).
Part B
(d) [1 mark]
A1: Correct calculation of \( I_0 = 1 \).
(e) [4 marks]
M1: Correct choice of parts \( u = x^n \) and \( dv = e^{-x} dx \).
A1: Correct application of integration by parts formula.
R1: Clear explanation of why the boundary term vanishes as \( x \to \infty \).
A1: Correctly obtaining the recurrence relation \( I_n = n I_{n-1} \).
(f) [1 mark]
A1: Deducing \( I_n = n! \).
(g) [5 marks]
A1: Correctly showing the base case \( n = 0 \).
M1: Stating assumption for \( n = k \) and attempting \( n = k+1 \) using integration by parts.
A1: Correct expression for \( J_{k+1}(t) \) in terms of \( J_k(t) \).
M1: Correct algebraic expansion and regrouping with the summation.
R1: Clear concluding statement indicating induction is complete.
Part C
(h) [1 mark]
A1: Correct sum expression for Poisson cumulative probability.
(i) [3 marks]
M1: Attempting to link \( J_n(t) \) and \( I_n \).
M1: Factoring \( n! \) and identifying the cumulative Poisson sum.
A1: Clearly showing the final integration identity.
(j) [3 marks]
M1: Identifying mean and variance of the Poisson distribution.
M1: Standardizing to the standard normal distribution.
A1: Evaluating to the limit value of 0.5.
(k) [3 marks]
M1: Linking the integral limit to the probability limit.
A2: Stating the final limit value is 0.5.