Chaitanya's Random Pages

March 28, 2014

Catching Fire vs Frozen

Filed under: movies and TV — ckrao @ 5:35 am

The movies Hunger Games: Catching Fire and Frozen have lit up the box office recently – worldwide they are the fifth and second biggest releases of 2013. Interestingly they were both released on the same day in the US and Canada (though in only one theatre in the case of Frozen) and the graph below shows the different rates at which they accumulated their totals.

catching fire vs frozen

Catching Fire followed a trajectory largely typical of large blockbusters: an opening weekend of $158m (6th largest ever unadjusted) leading to a gross around $425m corresponds to a multiplier of 2.7. It still made significant money during the Christmas break with a gross exceeding $10m in its sixth weekend which is rare. The course of Frozen started off similarly in terms of second weekend drop, but had a smaller opening weekend in wide release of $67m. Funnily they had an identical second weekend drop of 53.1% which impressive for both movies given the size of the first weekend for Catching Fire and the fact that it was post-Thanksgiving weekend for Frozen’s second weekend. Following Thanksgiving weekend, Catching Fire in fact had the 3rd largest gross after 10 days, behind only Avengers and The Dark Knight.

After 15 days Catching Fire had made $317m, Frozen was on $109m and both movies were making comparable amounts of money on a daily basis at that point. It is remarkable then that Frozen has ended up within $30m of Catching Fire’s total. Frozen had unbelievable staying power during the holiday season as seen from the above graph, with only one weekend drop exceeding 30% from its 3rd weekend until its release on Blu-ray/DVD (when still in the top ten!). Furthermore during that weekend (1) the drop was still just 31.5%, (2) it was after a boost the previous weekend due to the new year holidays, and (3) it was back up to number one movie at the box office! This type of behaviour is hardly ever seen in this age of front-loadedness (where movies usually make the bulk of their money in early weeks).

Frozen made more than half of its money in US + Canada after its fourth weekend of wide release and was in the top ten for an amazing 16 weeks. Its multiplier of close to 6 given that size of opening weekend is the best since Avatar and perhaps no other movie since Phantom Menace (released in 1999) has a comparable multiplier for a large opener. Nobody thought it would get close to $400m after being at $134m following a $31.6m second weekend! It has been fun following its box office journey. 🙂

March 26, 2014

Inverse variance weighting form of the conditional covariance of multivariate Gaussian vectors

Filed under: mathematics — ckrao @ 5:48 am

Let X and V be independent zero-mean real Gaussian vectors of respective length m and n with respective invertible covariance matrices \Sigma_X and \Sigma_V. Let C be a full-rank m \times n matrix and define Y by

\displaystyle Y = CX + V.\quad\quad(1)

If we are given the vector X we know that Y will be Gaussian with mean CX and covariance \Sigma_V.

However suppose we are given Y and wish to find the conditional distribution of X|Y. Here we may think of X as a hidden variable and Y as the observed variable. In this case the result is a little more involved. If we recover the results of this earlier blog post, (X^T,Y^T) is jointly Gaussian and so X|Y is Gaussian with mean

\displaystyle E[X|Y] = E[X] + \text{cov}(X,Y)(\text{cov}(Y))^{-1}(Y - E[Y])\quad\quad(2)

and covariance

\displaystyle \text{cov}(X|Y) = \text{cov}(X) - \text{cov}(X,Y)\text{cov}(Y)^{-1}\text{cov}(Y,X).\quad\quad(3)

(Here \text{cov}(A,B) := E[AB^T] is the cross-covariance of A and B, while \text{cov}(A):= \text{cov}(A,A).)

Using the fact that E(X) = 0, E(Y) = 0, \text{cov}(X) = \Sigma_X,

\text{cov}(X,Y) = E[X(CX+V)^T] = E[XX^T]C^T = \Sigma_X C^T

and \text{cov}(Y) = C\Sigma_X C^T + \Sigma_V, (2) and (3) become

\begin{aligned} E[X|Y] &= \Sigma_X C^T(C\Sigma_X C^T + \Sigma_V)^{-1}Y,\quad\quad&(4)\\ \text{cov}(X|Y) &= \Sigma_X - \Sigma_X C^T(C\Sigma_X C^T + \Sigma_V)^{-1}C\Sigma_x. \quad\quad&(5)\end{aligned}

In this post we also derive the following alternative expressions (also described here) which the covariances appear as inverse matrices.

\boxed{ \begin{aligned} E[X|Y] &= \Sigma C^T \Sigma_V^{-1} Y\quad\quad&(6)\\ \text{cov}(X|Y) &= \Sigma,\quad\quad&(7)\\\text{where}&&\\ \Sigma &:= (\Sigma_X^{-1} + C^T \Sigma_V^{-1} C)^{-1}.\quad\quad&(8)\end{aligned} }

Note that in the scalar case y = cx + v with variances \sigma_x^2 and \sigma_v^2 (5) and (7) become the identity

\displaystyle \sigma_x^2 - \frac{\sigma_x^4c^2}{c^2 \sigma_x^2 + \sigma_v^2} = (\sigma_x^{-2} + c^2\sigma_v^{-2})^{-1}.\quad\quad(9)

In the case where y is a scalar and C is a diagonal matrix, y is a weighted sum of the elements of vector X and (8) becomes the inverse of a sum of inverses of variances (inverse-variance weighting).

One can check algebraically that the expressions (4),(6) and (5),(7) are equivalent in the matrix case, or we may proceed as follows.

Let V = \left[ \begin{array}{cc} V_{11} & V_{12}\\ V_{21} & V_{22} \end{array} \right] = \left[ \begin{array}{cc}E(XX^T) & E(XY^T)\\ E(YX^T) & E(YY^T) \end{array} \right] be the covariance matrix of the joint vector \left[ \begin{array}{c} X\\ Y \end{array} \right]. Then since Y = CX + V we have V_{11} = \Sigma_X and

\begin{aligned}  V_{12} &= V_{21}^T\\  &= E(XY^T)\\  &= E(X(CX + V)^T\\  &= EXX^T C^T + EXV^T\\  &= \Sigma_X C^T. \quad\quad(10)  \end{aligned}

Then as the Gaussian vector \left[ \begin{array}{c} X\\ Y \end{array} \right] is zero-mean with covariance V, the joint pdf of X and Y is proportional to \exp \left(-\frac{1}{2} [X^T Y^T]V^{-1}\left[ \begin{array}{c} X\\ Y \end{array} \right] \right).

The key step now is to make use of the following identity (also see this explanation) based on completion of squares:

\displaystyle  \exp \left(-\frac{1}{2} [X^T Y^T]V^{-1}\left[ \begin{array}{c} X\\ Y \end{array} \right] \right)  = \exp\left( -\frac{1}{2} (X^T - Y^TA^T) S_{22}^{-1}(X-AY)\right) \exp\left( -\frac{1}{2} Y^T V_{22}^{-1} Y\right),\quad\quad(11)

where A and S_{22} are matrices, defined similarly to A and s in the scalar equation

\displaystyle ax^2 + 2bxy + cy^2 = a(x-Ay)^2 + sy^2.

We will show that A = V_{12}V_{22}^{-1} and S_{22} = V_{11} - V_{12}V_{22}^{-1}V_{21} (S_{22} is the Schur complement of V_{22} in V also discussed in this previous blog post).

The second term in the right side of (11) is proportional to the pdf of Y (being Gaussian) p(Y), so the first term must be proportional to the conditional pdf p(X|Y). We are left to find S_{22} = \text{cov}(X|Y) and AY = E(X|Y).

From (11), S_{22}^{-1} is the top-left block of V^{-1} while -S_{22}^{-1}A is the top-right block of V^{-1}. To find these blocks, consider the block matrix equation

\displaystyle V \left[ \begin{array}{c} r\\s \end{array} \right] = \left[ \begin{array}{c} a\\b \end{array} \right] \quad \Rightarrow \quad \left[ \begin{array}{c} r\\s \end{array} \right] = V^{-1} \left[ \begin{array}{c} a\\b \end{array} \right]. \quad\quad(12)

This is the same as the system of equations

\displaystyle \begin{aligned}  V_{11} r + V_{12} s &= a, \quad\quad&(13)\\  V_{21} r + V_{22} s &= b. \quad \quad&(14)\\  \end{aligned}

Multiplying (13) by V_{21}V_{11}^{-1} gives

\displaystyle V_{21}r + V_{21}V_{11}^{-1}V_{12}s = V_{21}V_{11}^{-1}a.

Subtracting this from (14) gives

\begin{aligned} (V_{22} - V_{21}V_{11}^{-1}V_{12})s &= b - V_{21}V_{11}^{-1}a\\ \Rightarrow s &= (V_{22} - V_{21}V_{11}^{-1}V_{12})^{-1}b - (V_{22} - V_{21}V_{11}^{-1}V_{12})V_{21}V_{11}^{-1}a.\quad \quad(15)\end{aligned}

Since \left[ \begin{array}{c} r\\s \end{array} \right] = V^{-1} \left[ \begin{array}{c} a\\b \end{array} \right] the coefficients of a and b in (15) are the bottom-left and bottom-right blocks of V^{-1} respectively. We may then write S_{11} := V_{22} - V_{21}V_{11}^{-1}V_{12} and so from (15)

\displaystyle S_{11} s = b - V_{21}V_{11}^{-1}a.\quad \quad(16)

Also we have from (13)

\displaystyle r = V_{11}^{-1} a - V_{11}^{-1} V_{12}s.\quad\quad (17)

Using (16) this becomes

\begin{aligned}  r &= V_{11}^{-1} a - V_{11}^{-1} V_{12} S_{11}^{-1}(b - V_{21}V_{11}^{-1}a)\\  &=(V_{11}^{-1} + V_{11}^{-1} V_{12} S_{11}^{-1} V_{21} V_{11}^{-1} )a - V_{11}^{-1} V_{12} S_{11}^{-1}b.\quad\quad(18)  \end{aligned}

Note that analogous to (16) we could have written

\displaystyle S_{22} r = a - V_{12}V_{22}^{-1} b \Rightarrow r = S_{22}^{-1} a -S_{22}^{-1}V_{12}V_{22}^{-1}b .\quad\quad(19)

Comparing coefficients of a, b of (18) and (19) gives

\displaystyle S_{22}^{-1} = V_{11}^{-1} + V_{11}^{-1} V_{12} S_{11}^{-1} V_{21} V_{11}^{-1}\quad\quad(20)


\displaystyle V_{11}^{-1} V_{12} S_{11}^{-1} = S_{22}^{-1} V_{12}V_{22}^{-1}.\quad\quad(21)

In the same way that S_{22} = \text{cov}(X|Y), S_{11} = \text{cov}(Y|X), which for Y = CX + V is simply \Sigma_V as we saw before.

Hence from (20),

\begin{aligned}  \Sigma^{-1} = S_{22}^{-1} &= V_{11}^{-1} + V_{11}^{-1} V_{12} S_{11}^{-1} V_{21} V_{11}^{-1}\\  &= \Sigma_X^{-1} + \Sigma_X^{-1}\Sigma_X C^T \Sigma_V^{-1} C \Sigma_X^{-1} \Sigma_X\\  &= \Sigma_X^{-1} + C^T \Sigma_V^{-1} C  \end{aligned}

and from the b-coefficient of (18),

\begin{aligned}  S_{22}^{-1}A &= V_{11}^{-1} V_{12} S_{11}^{-1}\\  \Rightarrow A &= S_{22} V_{11}^{-1} V_{12} S_{11}^{-1}\\  &= S_{22} \Sigma_X^{-1} \Sigma_X C^T \Sigma_V^{-1}\\  &= \Sigma C^T \Sigma_V^{-1}.  \end{aligned}

Hence E[X|Y] = AY = \Sigma C^T \Sigma_V^{-1} Y as desired, and equations (6)-(8) have been verified.

Create a free website or blog at

%d bloggers like this: