### Why Cable Companies Bundle

**Matthew Martin**1/27/2013 05:00:00 AM

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There's been a bit of debate on why cable companies bundle channels, rather than charge subscribers a per-channel fee. Brian Stelter argues that we are all implicitly subsidizing sports fans, while Matt Yglesias argues that the issue is lack of competition, and that a per-channel price structure is unlikely to change what you pay.Warning:this post is a bit mathematical. I have written what I think should be a more intuitive, non-math version of this post here.

Yglesias is right that competition (lack thereof) is of critical importance. And, short of having the government administer cable lines the same way they do highways, it is hard to see how we can get competition in this market. However, he is wrong to suggest that bundling is no worse than a per-channel price structure. Here's the model:

The cable provider provides three channels--$ABC$, $CBS$, and $NBC$--and has a complete monopoly. However, any offer made to one of its subscribers is available to all subscribers--the company cannot simply charge people different prices for the same bundles. There are two types of consumers, the $H$ type are avid watchers with a high value from television, while the $L$ type derive low value from television. Lets suppose that there are two $H$ types and only one $L$ type. In addition to the three TV stations, there is a composite commodity $X$ representing all other goods and services that the consumers want (which are not provided by the cable company). The utility of consumer $H$ is $$ U_H=ln\left(X\right)+ln\left(ABC+CBS+NBC+1\right)$$ while the utility of consumer $L$ is $$ U_L=ln\left(X\right)+\frac{1}{2}ln\left(ABC+CBS+NBC+1\right) $$ Each consumer has wealth $M$ that he can spend on either TV or other stuff $X$.

If the cable company could, it would like to provide all three channels to all three consumers, but charge $P_L^*=\frac{M}{2}$ to $L$ type consumers and $P_H^*=\frac{3}{4}M$ to $H$ type consumers (See Appendix [A] after the jump below), but by law the offer $P_L$ is available to all $H$ types as well, so all three consumers will pay $P_L$ instead, for a profit to the cable company of $\pi^*=\frac{3}{2}M$, which is the same as if the firm charges $P_H^*$, which only $H$ types will buy.

The cable company can make more profits by offering two types bundles: paying $B_1$ would get you only one of the channels (for our purposes, it doesn't matter which one), while paying $B_3$ gets you all three channels. It turns out (derived below) that the profit maximizing prices are $B_1^*=\frac{\sqrt{2}-1}{\sqrt{2}}M$ and $B_3^*= \frac{4-\sqrt{2}}{4}M$. Consumer $L$ will buy $B_1$ while consumer $H$ will buy $B_3$, which yields profits $\pi_{1,3}^*=B_1^*+2B_3^*$, which is larger than above. This yields higher profits than any other possible price/bundle scheme. (See appendix [B]).

In particular, this is more profitable than charging individual prices for each network. Suppose now that the cable company must choose prices $P_1$ for $ABC$, $P_2$ for $NBC$, and $P_3$ for $CBS$. We're going to solve this profit-maximization problem with pure brute force. First, let us define the prices such that $P_1\leq P_2 \leq P_2 \leq P_3$, which is merely a matter of notation.

Lets consider first the high types only. To get $H$ to buy at least one channel, we must have that $P_1\leq\frac{M}{2}$. To get $H$ to buy a second channel, we must have that $P_2 \leq \frac{M}{3}-\frac{P_1}{3}$. To get the household to buy a third channel, we must have $P_3\leq \frac{1}{4}\left(M-P_2-P_1\right)$ Together with the constraint that $P_1\leq P_2 \leq P_3$ , we have a constraint set describing the possible ways to induce the $H$ types to buy one, two, or three channels. Note that the constraints are nested, so that it must be the case that all three prices must be equal to each other. This is not a surprising result since both types of consumers regard the three channels as perfect substitutes. (Appendix [C])

We do the same for low types. To get $L$ to buy one channel we need to charge $P_1\leq\left(1-\frac{1}{\sqrt{2}}\right)M$, to get him to buy a second we need to charge $P_2\leq\left(1-\sqrt{\frac{2}{3}}\right) \left(M-P_1\right) $, and for the third we need to charge $P_3\leq\left(1-\sqrt{\frac{3}{4}}\right) \left(M-P_1-P_2\right) $. In addition to the constraint $P_1\leq P_2 \leq P_3$ this defines a set of possible price strategies, as before. We will compute the profit-maximizing strategy by assembling a table (These are coefficients--you need to multiply the values by $M$ to get the actual prices and profits):

Channels | L | Profits | H | Profits |

1 | 0.292893 | 0.292893 | 0.5 | 1 |

2 | 0.155051 | 1.240408 | 0.25 | 1.25 |

3 | 0.062782 | 0.565035 | 0.166667 | 1.166667 |

The profits from bundling are $\pi_{1,3}^*= 1.585786$ which is higher than any of the strategies in the table. Thus, bundling is more profitable--and more expensive to consumers--than mandating a per-channel price and allowing consumers to choose which channels they want. And, as shown above, this is even true when we allow different prices for different channels. And the difference is dramatic--if we banned bundling, more people would have more channels at a fraction of the cost.

Proofs and Derivations after the jump:

[A] We compute the maximum amount that each type of household would be willing to pay for all three channels by calculating the maximum prices $P_L,~P_H$ such that utility is higher than not buying any TV. For $H$ we need the maximum $P_H$ such that: $$ln\left(M\right)\leq ln\left(M-P_H\right)+ln\left(4\right)$$ and similarly for $L$ we solve for the maximum $P_L$ such that $$ ln\left(M\right)\leq ln\left(M-P_L\right)+\frac{1}{2}ln\left(4\right)$$

[B] $B_1^*$ is the maximum amount that household $L$ is willing to pay for a single channel. That is, we solve for the maximum $B_1$ for which $$ ln\left(M\right)\leq ln\left(M-B_1\right)+\frac{1}{2}ln\left(2\right)$$ Similarly, $B_3^*$ is the maximum amount household type $H$ is willing to pay to get the two additional channels not available with $B_1^*$. We solve for the maximum $B_3$ for which $$ ln\left(M-B_1^*\right)+ln\left(2\right)\leq ln\left(M-B_3\right)+ln\left(4\right)$$ Simply adding revenues over two type $H$ households and one type $L$ household shows this is more profitable than selling bundles of all three stations to both types at price $P_L^*$. Since the cable company faces no costs of its own, we know that it can always get more profit pricing bundles so that at least one household will buy all three channels. Moreover, the $H$ type households have a higher value from TV, so we know they will definitely buy more channels than the $L$ types. Hence, it is sufficient to check that this menu is more profitable than one which includes an option for household $L$ to buy a bundle $B_2$ of two channels. We solve for the maximum$B_2$ that $L$ will pay for two bundles $$ ln\left(M\right)\leq ln\left(M-B_2\right)+\frac{1}{2}ln\left(3\right)$$ and note that at this price $B_2^*$ the $H$ type households would prefer to buy the two-channel option to $B_3^*$: $$ln\left(M-B_2^*\right)+ln\left(3\right) > ln\left(M-B_2^*\right)+ln\left(4\right) $$ Finally, we verify that $3B_2^*<\pi_{1,3}^*$

[C] We have ordered the prices of each channel in order from lowest to highest. The logic here is that since the households are indifferent between channels, they will buy the cheapest one first. That is, if the household is paying $P_2$ then it has already paid $P_1$ for their first channel, and if the household is paying $P_3$, then it has already paid $P_2$ for the second channel and $P_1$ for the first channel. We have three inequalities establishing maximum bounds on each of the prices in terms of what they previously paid--if the price is higher than this, they won't buy the extra channel. Moreover, the three sets are nested--the set of possible $P_3$ is contained by the set of possible $P_2$ which is contained by the set of possible $P_1$. Profit maximization dictates we choose the maximum possible $P_3$ for given $P_1,P_2$, which means that $P_3=\frac{1}{4}\left(M-P_1-P_2\right)$, and the maximum possible $P_2$ given $P_1$ which means $P_2= \frac{1}{3}\left(M-P_1\right)$ . Finally, we note that given these constraints, profits are increasing in $P_1$. Together with the fact that, by construction, $P_1\leq P_2\leq P_3$ this proves that $P_3=P_2=P_1$. We use this fact to solve for their values, recorded in the table shown above, under six possible scenarios: where the $L$ types buy one, two, and three channels respectively in column two, and where the $H$ types buy one, two, and three channels respectively in column four.